RS Aggarwal Solutions for Class 12 Chapter 32 Binomial Distribution

Question 1. A coin is tossed 6 times. Find the probability of getting at least 3 heads.
Answer: When a coin is tossed 6 times, the total outcomes = 2^6 = 64. The favorable outcomes for getting at least 3 heads are: 6C₃ + 6C₄ + 6C₅ + 6C₆. Using the binomial coefficients: (20 + 15 + 6 + 1) = 42. So the probability = 42/64 = 21/32.
In simple words: Toss the coin 6 times and count how many times you get 3 or more heads. Out of all possible results, 21 out of 32 will show at least 3 heads.

Exam Tip: Remember that "at least 3" includes exactly 3, exactly 4, exactly 5, and exactly 6 heads - add all these cases together.

Question 2. A coin is tossed 5 times. What is the probability that a head appears an even number of times?
Answer: When a coin is tossed 5 times, the total outcomes = 2^5 = 32. For a head appearing an even number of times, we need either 0, 2, or 4 heads. The favorable outcomes are: 5C₀ + 5C₂ + 5C₄ = (1 + 10 + 5) = 16. Therefore, the probability = 16/32 = 1/2.
In simple words: An even number means 0, 2, or 4 heads - never 1, 3, or 5. When you add up all the ways this can happen, exactly half of all possible outcomes will have an even number of heads.

Exam Tip: Always identify what "even" means in the problem context - here it refers to the count of heads, not the number of trials.

Question 3. 7 coins are tossed simultaneously. What is the probability that a tail appears an odd number of times?
Answer: When 7 coins are tossed together, the total outcomes = 2^7 = 128. For a tail appearing an odd number of times, we need 1, 3, 5, or 7 tails. The favorable outcomes are: 7C₁ + 7C₃ + 7C₅ + 7C₇ = (7 + 35 + 21 + 1) = 64. Thus, the probability = 64/128 = 1/2.
In simple words: Out of every 2 sets of results, one set will have an odd number of tails (1, 3, 5, or 7 tails), and the other will have an even number.

Exam Tip: Notice that "odd number of tails" is equally likely as "even number of tails" - this is a general rule for coin tosses.

Question 4. A coin is tossed 6 times. Find the probability of getting
(i) exactly 4 heads
(ii) at least 1 heads
(iii) at most 4 heads
Answer:
(i) When the coin is tossed 6 times, total outcomes = 2^6 = 64. The favorable outcomes for exactly 4 heads = 6C₄ = 15. Therefore, probability = 15/64.
(ii) Total outcomes = 64. The favorable outcomes for at least 1 head = 6C₁ + 6C₂ + 6C₃ + 6C₄ + 6C₅ + 6C₆ = (6 + 15 + 20 + 15 + 6 + 1) = 63. Therefore, probability = 63/64.
(iii) Total outcomes = 64. The favorable outcomes for at most 4 heads = 6C₀ + 6C₁ + 6C₂ + 6C₃ + 6C₄ = (1 + 6 + 15 + 20 + 15) = 57. Therefore, probability = 57/64.
In simple words: (i) Getting exactly 4 heads is fairly common - about 15 out of 64 ways. (ii) Getting at least 1 head is almost certain - only 1 way out of 64 gives no heads at all. (iii) Getting at most 4 heads means we accept 0, 1, 2, 3, or 4 heads.

Exam Tip: "At least 1" is easier to calculate as 1 minus "exactly 0". Similarly, "at most 4" = 1 minus "at least 5".

Question 5. 10 coins are tossed simultaneously. Find the probability of getting
(i) exactly 3 heads
(ii) not more than 4 heads
(iii) at least 4 heads
Answer:
(i) When 10 coins are tossed, total outcomes = 2^10 = 1024. The favorable outcomes for exactly 3 heads = 10C₃ = 120. Therefore, probability = 120/1024 = 15/128.
(ii) Total outcomes = 1024. The favorable outcomes for not more than 4 heads = 10C₀ + 10C₁ + 10C₂ + 10C₃ + 10C₄ = (1 + 10 + 45 + 120 + 210) = 386. Therefore, probability = 386/1024 = 193/512.
(iii) Total outcomes = 1024. The favorable outcomes for at least 4 heads = 10C₄ + 10C₅ + 10C₆ + 10C₇ + 10C₈ + 10C₉ + 10C₁₀ = 848. Therefore, probability = 848/1024 = 53/64.
In simple words: (i) With 10 coins, getting exactly 3 heads is one specific outcome among many. (ii) Getting 4 or fewer heads means most coins show tails. (iii) Getting 4 or more heads means at least some coins show heads.

Exam Tip: Always simplify final fractions by finding the greatest common divisor before writing your answer.

Question 6. A die is thrown 6 times. If 'getting an even number' is a success, find the probability of getting
(i) exactly 5 successes
(ii) at least 5 successes
(iii) at most 5 successes
Answer:
(i) Using Bernoulli's Trial formula P(X = x) = ⁿCₓ·p^x·q^(n-x), where n = 6, success = rolling 2, 4, or 6, so p = 1/2 and q = 1/2. For exactly 5 successes: P(X = 5) = 6C₅·(1/2)^5·(1/2)^1 = 6·(1/64) = 6/64 = 3/32.
(ii) For at least 5 successes: P(X = 5) + P(X = 6) = [6C₅·(1/2)^5·(1/2)^1] + [6C₆·(1/2)^6·(1/2)^0] = (6/64) + (1/64) = 7/64.
(iii) For at most 5 successes: This equals 1 - P(X = 6) = 1 - (1/64) = 63/64.
In simple words: (i) When rolling the die 6 times, getting exactly 5 even numbers happens 3 times out of 32 tries. (ii) Getting 5 or 6 even numbers is quite rare. (iii) Getting 5 or fewer even numbers is almost certain - you'd need all 6 rolls to be odd for this to fail.

Exam Tip: Identify success and failure probabilities first - for a die, even numbers (2, 4, 6) give p = 1/2.

Question 7. A die is thrown 4 times. 'Getting a 1 or a 6' is considered a success. Find the probability of getting
(i) exactly 3 successes
(ii) at least 2 successes
(iii) at most 2 successes
Answer:
(i) Using Bernoulli's Trial, n = 4, p = 2/6 = 1/3 (rolling 1 or 6), q = 4/6 = 2/3. For exactly 3 successes: P(X = 3) = 4C₃·(1/3)^3·(2/3)^1 = 4·(1/27)·(2/3) = 8/81.
(ii) For at least 2 successes: P(X ≥ 2) = P(X = 2) + P(X = 3) + P(X = 4). Calculate: 4C₂·(1/3)^2·(2/3)^2 + 4C₃·(1/3)^3·(2/3)^1 + 4C₄·(1/3)^4·(2/3)^0 = (24/81) + (8/81) + (1/81) = 33/81 = 11/27.
(iii) For at most 2 successes: P(X ≤ 2) = P(X = 0) + P(X = 1) + P(X = 2) = (16/81) + (32/81) + (24/81) = 72/81 = 8/9.
In simple words: (i) Getting exactly 3 rolls that show 1 or 6 happens in 8 out of 81 cases. (ii) Getting 2 or more such rolls happens fairly often - about 11 out of 27 times. (iii) Getting 2 or fewer such rolls is even more likely - about 8 out of 9 times.

Exam Tip: When defining success, count all favorable outcomes carefully - here, success means rolling either 1 or 6, so p = 2/6 = 1/3.

Question 8. Find the probability of a 4 turning up at least once in two tosses of a fair die.
Answer: When tossing a die twice, the total possible outcomes = 36. The favorable outcomes where at least one 4 appears are: (1,4), (2,4), (3,4), (4,4), (5,4), (6,4), (4,1), (4,2), (4,3), (4,5), (4,6), which gives 11 outcomes. Therefore, the probability = 11/36.
In simple words: Out of all 36 ways two dice can land, 11 of them will show at least one 4. It's easier to see this by noting: 1 way both show 4, and 10 ways exactly one shows 4.

Exam Tip: For "at least once" problems, often it's simpler to calculate 1 minus the probability of "never" - here, P(no 4) = (5/6)^2 = 25/36, so P(at least one 4) = 1 - 25/36 = 11/36.

Question 9. A pair of dice is thrown 4 times. If 'getting a doublet' is considered a success, find the probability of getting 2 successes.
Answer: A pair of dice has 36 total outcomes. A doublet (both dice show the same number) occurs in 6 ways: (1,1), (2,2), (3,3), (4,4), (5,5), (6,6). So p = 6/36 = 1/6 and q = 30/36 = 5/6. Using Bernoulli's Trial with n = 4, x = 2: P(X = 2) = 4C₂·(1/6)^2·(5/6)^2 = 6·(1/36)·(25/36) = 150/1296 = 25/216.
In simple words: When throwing two dice 4 times, the chance of getting exactly 2 doublets (pairs like 2-2 or 5-5) is 25 out of 216 tries.

Exam Tip: Identify the success event clearly - a doublet means both dice show identical numbers, which happens 6 times out of 36 possible outcomes.

Question 10. A pair of dice is thrown 7 times. If 'getting a total of 7' is considered a success, find the probability of getting
(i) no success
(ii) exactly 6 successes
(iii) at least 6 successes
(iv) at most 6 successes
Answer:
(i) Getting a total of 7 occurs with pairs: (1,6), (6,1), (2,5), (5,2), (3,4), (4,3) - that's 6 ways out of 36. So p = 6/36 = 1/6, q = 5/6. For no success in 7 throws: P(X = 0) = 7C₀·(1/6)^0·(5/6)^7 = (5/6)^7.
(ii) For exactly 6 successes: P(X = 6) = 7C₆·(1/6)^6·(5/6)^1 = 7·(1/6)^6·(5/6) = 35·(1/6)^7.
(iii) For at least 6 successes: P(X = 6) + P(X = 7) = [7C₆·(1/6)^6·(5/6)^1] + [7C₇·(1/6)^7·(5/6)^0] = [35·(1/6)^7] + [(1/6)^7] = 36·(1/6)^7 = (1/6)^5 when simplified.
(iv) For at most 6 successes: This equals 1 - P(X = 7) = 1 - (1/6)^7.
In simple words: (i) Getting no 7-totals in 7 throws is quite likely. (ii) Getting exactly 6 is extremely rare. (iii) Getting 6 or more is almost impossible. (iv) Getting 6 or fewer is nearly certain since getting all seven 7-totals is virtually impossible.

Exam Tip: Always list the pairs that create your "success" outcome first - for a sum of 7, there are exactly 6 favorable pairs.

Question 11. There are 6% defective items in a large bulk of times. Find the probability that a sample of 8 items will include not more than one detective item.
Answer: Using Bernoulli's Trial with n = 8, the probability of defective items p = 6/100 = 0.06, and q = 94/100 = 0.94. For not more than one defective item, we calculate: P(X = 0) + P(X = 1). This equals 8C₀·(0.06)^0·(0.94)^8 + 8C₁·(0.06)^1·(0.94)^7 = (0.94)^8 + 8·(0.06)·(0.94)^7 = (0.94)^7·[(0.94) + (0.48)] = (0.94)^7·(1.42).
In simple words: When picking 8 items from a batch where 6% are faulty, the chance of getting either no faulty items or exactly 1 faulty item is quite high because faults are rare.

Exam Tip: Convert percentages to decimals early - here 6% becomes 0.06 for easier calculation with the binomial formula.

Question 12. In a box containing 60 bulbs, 6 are defective. What is the probability that out of a sample of 5 bulbs
(i) none is defective
(ii) exactly 2 are defective
Answer:
(i) The probability of selecting a defective bulb p = 6/60 = 1/10, q = 54/60 = 9/10. Using Bernoulli's Trial with n = 5, for no defective bulbs: P(X = 0) = 5C₀·(1/10)^0·(9/10)^5 = (9/10)^5.
(ii) For exactly 2 defective bulbs: P(X = 2) = 5C₂·(1/10)^2·(9/10)^3 = 10·(1/100)·(729/1000) = 7290/100000 = 729/10000.
In simple words: (i) The chance that all 5 bulbs picked are good is (9/10)^5, which is pretty likely since only 1 in 10 are faulty. (ii) The chance of exactly 2 faulty bulbs is much smaller.

Exam Tip: Simplify the probability of success first - here 6/60 reduces to 1/10, making calculations cleaner.

Question 13. The probability that a bulb produced by a factory will fuse after 6 months of use is 0.05. Find the probability that out of 5 such bulbs
(i) none will fuse after 6 months of use
(ii) at least one will fuse after 6 months of use
(iii) not more than one will fuse after 6 months of use
Answer:
(i) With p = 0.05 and q = 0.95, for none fusing: P(X = 0) = 5C₀·(0.05)^0·(0.95)^5 = (0.95)^5.
(ii) For at least one fusing: P(X ≥ 1) = 1 - P(X = 0) = 1 - (0.95)^5.
(iii) For not more than one fusing: P(X ≤ 1) = P(X = 0) + P(X = 1) = (0.95)^5 + 5·(0.05)·(0.95)^4 = (0.95)^4·[(0.95) + (0.25)] = (0.95)^4·(1.20).
In simple words: (i) Since only 5% of bulbs fuse, getting 5 that all work is fairly likely. (ii) Getting at least one that fuses is the complement - less likely. (iii) Getting 0 or 1 fused is the most probable outcome.

Exam Tip: When p is already given as a decimal, use it directly in the binomial formula rather than converting.

Question 14. In the items produced by a factory, there are 10% defective items. A sample of 6 items is randomly chosen. Find the probability that this sample contains
(i) exactly 2 defective items
(ii) not more than 2 defective items
(iii) at least 3 defective items
Answer:
(i) With p = 1/10 and q = 9/10, for exactly 2 defective items: P(X = 2) = 6C₂·(1/10)^2·(9/10)^4 = 15·(1/100)·(6561/10000) = 98415/1000000. Simplifying: 3·(9/10)^4/20.
(ii) For not more than 2 defective: P(X ≤ 2) = P(X = 0) + P(X = 1) + P(X = 2) = (9/10)^6 + 6·(1/10)·(9/10)^5 + 15·(1/10)^2·(9/10)^4.
(iii) For at least 3 defective: P(X ≥ 3) = 1 - P(X ≤ 2).
In simple words: (i) Getting exactly 2 faulty items out of 6 is somewhat possible but not very likely. (ii) Getting 2 or fewer faulty items is quite probable. (iii) Getting 3 or more is much less likely since only 10% are defective overall.

Exam Tip: For "at least" problems, sometimes it's easier to compute 1 minus the complement rather than summing many terms.

Question 15. Assume that on an average one telephone number out of 15, called between 3 p.m. on weekdays, will be busy. What is the probability that if six randomly selected telephone numbers are called, at least 3 of them will be busy?
Answer: The probability that a number is busy is p = 1/15, and q = 14/15. Using Bernoulli's Trial with n = 6, for at least 3 busy: P(X ≥ 3) = 1 - [P(X = 0) + P(X = 1) + P(X = 2)]. This equals 1 - [6C₀·(1/15)^0·(14/15)^6 + 6C₁·(1/15)^1·(14/15)^5 + 6C₂·(1/15)^2·(14/15)^4] = 1 - (14/15)^4·[(14/15)^2 + 6·(1/15)·(14/15) + 15·(1/15)^2].
In simple words: Since numbers are busy only 1 in 15 times, getting 3 or more busy lines out of 6 calls is very rare and unlikely to happen.

Exam Tip: When success probability is very small (like 1/15), "at least" problems are best solved using the complement approach.

Question 16. Three cars participate in a race. The probability that any one of them has an accident is 0.1. Find the probability that all the cars reach the finishing line without any accident.
Answer: If the probability of having an accident is 0.1, then the probability of reaching safely without accident is 1 - 0.1 = 0.9. Since the three cars race independently, the probability that all three reach safely = (0.9) × (0.9) × (0.9) = (0.9)^3 = 0.729.
In simple words: Each car has a 0.9 chance of finishing safely. Since all three must finish safely, multiply their individual probabilities together to get 0.729.

Exam Tip: For independent events, multiply individual probabilities. Here, "all three reach safely" requires all three conditions to occur simultaneously.

Question 17. Past records show that 80% of the operations performed by a certain doctor were successful. If the doctor performs 4 operations in a day, what is the probability that at least 3 operations will be successful?
Answer: With success probability p = 0.8 and q = 0.2, using Bernoulli's Trial for at least 3 successes out of 4 operations: P(X ≥ 3) = P(X = 3) + P(X = 4) = 4C₃·(0.8)^3·(0.2)^1 + 4C₄·(0.8)^4·(0.2)^0 = 4·(0.512)·(0.2) + (0.4096) = 0.4096 + 0.4096 = 0.8192 = 8192/10000 = 1024/1250 = 512/625.
In simple words: Since the doctor succeeds 80% of the time, getting 3 or 4 successes out of 4 operations is quite likely - about 51% chance.

Exam Tip: When success rate is high (like 80%), "at least" many successes becomes more probable.

Question 18. The probability of a man hitting a target is (1/4). If he fires 7 times, what is the probability of his hitting the target at least twice?
Answer: With p = 1/4 and q = 3/4, for at least 2 hits out of 7 shots: P(X ≥ 2) = 1 - P(X < 2) = 1 - [P(X = 0) + P(X = 1)] = 1 - [7C₀·(1/4)^0·(3/4)^7 + 7C₁·(1/4)^1·(3/4)^6] = 1 - [(3/4)^7 + 7·(1/4)·(3/4)^6] = 1 - (3/4)^6·[(3/4) + (7/4)].
In simple words: The marksman hits 1 out of 4 times on average. Getting at least 2 hits in 7 tries is fairly probable since he gets 7 chances.

Exam Tip: Use the complement (1 minus the opposite) when calculating "at least" probabilities with small p values and large n.

Question 19. In a hurdles race, a player has to cross 10 hurdles. The probability that he will clear each hurdle is (5/6). What is the probability that he will knock down fewer than 2 hurdles?
Answer: If clearing probability is 5/6, then knocking down probability q = 1/6. For knocking down fewer than 2 (meaning 0 or 1), with n = 10: P(X < 2) = P(X = 0) + P(X = 1) = 10C₀·(1/6)^0·(5/6)^10 + 10C₁·(1/6)^1·(5/6)^9 = (5/6)^10 + 10·(1/6)·(5/6)^9 = (5/6)^9·[(5/6) + (10/6)] = (5/6)^9·(15/6).
In simple words: The athlete clears most hurdles easily (5 out of 6 times), so the chance of knocking down fewer than 2 is quite high.

Exam Tip: When success probability is high, few failures occur, so "fewer than 2 failures" becomes very likely.

Question 20. A man can hit a bird, once in 3 shots. On this assumption he fires 3 shots. What is the chance that at least one bird is hit?
Answer: With hit probability p = 1/3 and q = 2/3, for at least 1 hit out of 3 shots: P(X ≥ 1) = 1 - P(X = 0) = 1 - 3C₀·(1/3)^0·(2/3)^3 = 1 - (2/3)^3 = 1 - 8/27 = 19/27.
In simple words: The shooter hits 1 out of 3 times. With 3 shots, the chance of getting at least one hit is 19 out of 27.

Exam Tip: Always use the complement when "at least 1" is requested - calculate 1 minus "zero successes" for simplicity.

Question 21. If the probability that a man aged 60 will live to be 70 is 0.65, what is the probability that out of 10 men, now 60, at least 8 will live to be 70?
Answer: With p = 0.65 and q = 0.35, for at least 8 men living out of 10 to age 70: P(X ≥ 8) = P(X = 8) + P(X = 9) + P(X = 10) = 10C₈·(0.65)^8·(0.35)^2 + 10C₉·(0.65)^9·(0.35)^1 + 10C₁₀·(0.65)^10·(0.35)^0. Computing these terms gives approximately 0.2615.
In simple words: Since 65% of men this age survive 10 years, getting 8 or more out of 10 to survive is somewhat likely but not certain.

Exam Tip: For "at least" with higher values and moderate p, sum the individual probability terms rather than using the complement.

Question 22. A bag contains 5 white, 7 red 8 black balls. If four balls are drawn one by one with replacement, what is the probability that
(i) None is white
(ii) All are white
(iii) At least one is white
Answer:
(i) Total balls = 5 + 7 + 8 = 20. Probability of non-white = 15/20 = 3/4. Drawing 4 balls with replacement, none white: P = (3/4)^4 = 81/256.
(ii) Probability of white = 5/20 = 1/4. All 4 white: P = (1/4)^4 = 1/256.
(iii) At least one white: P = 1 - P(none white) = 1 - 81/256 = 175/256.
In simple words: (i) Most balls are not white, so getting none white is fairly likely. (ii) Getting all white is very rare. (iii) Getting at least one white is much more probable than none.

Exam Tip: With replacement, each draw is independent, so multiply probabilities for multiple trials.

Question 23. A policeman fires 6 bullets at a burglar. The probability that the burglar will be hit by a bullet is 0.6. What is the probability that burglar is still unhurt?
Answer: If the hit probability is 0.6, then miss probability q = 0.4. For the burglar to remain unhurt (all 6 bullets miss): P(X = 0) = 6C₀·(0.6)^0·(0.4)^6 = (0.4)^6 = 0.004096.
In simple words: With a 60% hit chance per bullet, getting all 6 bullets to miss is extremely unlikely - only about 4 in a thousand.

Exam Tip: "Unhurt" means zero successes (zero hits), which is the lowest possible outcome in this scenario.

Question 24. A die is tossed thrice. A success is 1 or 6 on a toss. Find the mean and variance of successes.
Answer: With p = 2/6 = 1/3 (rolling 1 or 6) and q = 2/3, the probability distribution for 3 tosses is:
P(X = 0) = 3C₀·(1/3)^0·(2/3)^3 = 8/27
P(X = 1) = 3C₁·(1/3)^1·(2/3)^2 = 12/27
P(X = 2) = 3C₂·(1/3)^2·(2/3)^1 = 6/27
P(X = 3) = 3C₃·(1/3)^3·(2/3)^0 = 1/27

Mean μ = Σ(x·P(x)) = 0·(8/27) + 1·(12/27) + 2·(6/27) + 3·(1/27) = (12 + 12 + 3)/27 = 27/27 = 1

For variance, calculate E(X^2) = Σ(x²·P(x)) = 0·(8/27) + 1·(12/27) + 4·(6/27) + 9·(1/27) = (12 + 24 + 9)/27 = 45/27 = 5/3

Variance σ² = E(X²) - μ² = 5/3 - 1 = 2/3
In simple words: On average, you get 1 success (rolling 1 or 6) when tossing the die 3 times. The variance of 2/3 shows how much the actual results vary around this average.

Exam Tip: For a binomial distribution, mean = np and variance = npq - these formulas provide quick checks of your calculations.

Question 25. A die is thrown 100 times. Getting an even number is considered a success. Find the mean and variance of success.
Answer: The likelihood of rolling an even number equals 3/6 = 1/2. The likelihood of rolling an odd number equals 3/6 = 1/2. Using the variance formula \( \text{Variance} = npq \), we get \( 100 \times \frac{1}{2} \times \frac{1}{2} = 25 \).
In simple words: When you roll the die 100 times, the average number of even rolls expected is 50, and how much those results will vary is 25.

Exam Tip: Always remember that for a binomial distribution, variance = npq, where n is trials, p is success probability, and q is failure probability.

Question 26. Determine the binomial distribution whose mean is 9 and variance is 6.
Answer: Given Mean = np = 9 and Variance = npq = 6, we find q = 6/9 = 2/3. Then p = 1 - 2/3 = 1/3. Solving for n: n = 9 / (1/3) = 27. The binomial distribution is \( ^{27}C_r \left( \frac{1}{3} \right)^r \left( \frac{2}{3} \right)^{27-r} \) where r = 0, 1, 2, 3, …, 27.
In simple words: We work backward from the mean and variance to find that we need 27 trials, with success probability 1/3 and failure probability 2/3.

Exam Tip: Use the relationship between mean, variance, and the parameters n, p, q to set up a system of equations - this is a key technique for finding the distribution.

Question 27. Find the binomial distribution whose mean is 5 and variance is 2.5.
Answer: From Mean = np = 5 and Variance = npq = 2.5, we calculate q = 2.5/5 = 1/2. Therefore p = 1 - 1/2 = 1/2. We find n = 5 / (1/2) = 10. The probability distribution is \( ^{10}C_r \left( \frac{1}{2} \right)^r \left( \frac{1}{2} \right)^{10-r} \) where 0 ≤ r ≤ 10.
In simple words: The distribution has 10 trials with both success and failure having equal probability of 1/2 each.

Exam Tip: When variance equals half the mean, this suggests equal probabilities for success and failure (p = q = 1/2).

Question 28. The mean and variance of a binomial distribution are 4 and (4/3) respectively. Find P(X ≥ 1).
Answer: From Mean = np = 4 and Variance = npq = 4/3, we get q = (4/3)/4 = 1/3. Thus p = 1 - 1/3 = 2/3 and n = 4 / (2/3) = 6. To find P(X ≥ 1), we use P(X ≥ 1) = 1 - P(X = 0) = 1 - \( ^6C_0 \left( \frac{2}{3} \right)^0 \left( \frac{1}{3} \right)^6 \) = 1 - 1/729 = 728/729.
In simple words: It is easier to calculate the complement - find the chance of zero successes and subtract from 1 to get the chance of at least one success.

Exam Tip: Using the complement rule P(X ≥ 1) = 1 - P(X = 0) simplifies calculations significantly in binomial problems.

Question 29. For a binomial distribution, the mean is 6 and the standard deviation is \( \sqrt{2} \). Find the probability of getting 5 successes.
Answer: From Mean = np = 6 and Standard deviation = \( \sqrt{2} \), the variance = 2. Thus npq = 2, which gives q = 2/6 = 1/3 and p = 1 - 1/3 = 2/3. We find n = 6 / (2/3) = 9. The probability of getting 5 successes is \( P(X = 5) = ^9C_5 \left( \frac{2}{3} \right)^5 \left( \frac{1}{3} \right)^4 \).
In simple words: First extract n, p, and q from the given mean and standard deviation, then use the binomial formula to find the specific probability.

Exam Tip: Remember that standard deviation is the square root of variance - always square it first when solving problems.

Question 30. In a binomial distribution, the sum and the product of the mean and the variance are (25/3) and (50/3) respectively. Find the distribution.
Answer: Let Mean = np and Variance = npq. We have Mean + Variance = np + npq = np(1 + q) = 25/3 and (Mean)(Variance) = (np)(npq) = \( n^2p^2q = \frac{50}{3} \). From these equations, after dividing to eliminate n and solving the resulting quadratic, we find \( 6q^2 - 13q + 6 = 0 \). Solving gives q = 2/3 or q = 3/2. Since q cannot exceed 1, we take q = 2/3, so p = 1/3 and n = 15. The binomial distribution is \( ^{15}C_r \left( \frac{1}{3} \right)^r \left( \frac{2}{3} \right)^{15-r} \).
In simple words: Two conditions about the mean and variance give us equations that we solve to find p, q, and n, then write out the full distribution.

Exam Tip: Always reject probability values greater than 1 or less than 0 - this constraint is essential for validity.

Question 31. Obtain the binomial distribution whose mean is 10 and standard deviation is \( 2\sqrt{2} \).
Answer: From Mean = np = 10 and Standard deviation = \( 2\sqrt{2} \), the variance = \( (2\sqrt{2})^2 = 8 \). So npq = 8, giving q = 8/10 = 4/5 and p = 1 - 4/5 = 1/5. We find n = 10 / (1/5) = 50. The binomial distribution is \( ^{50}C_r \left( \frac{1}{5} \right)^r \left( \frac{4}{5} \right)^{50-r} \) where 0 ≤ r ≤ 50.
In simple words: Convert standard deviation to variance by squaring, then use the mean and variance to find all parameters of the distribution.

Exam Tip: Always square the standard deviation to get variance - this is a frequent source of calculation errors.

Question 32. Bring out the fallacy, if any, in the following statement: 'The mean of a binomial distribution is 6 and its variance is 9'
Answer: The statement contains a logical impossibility. For a valid binomial distribution, variance must always be less than or equal to the mean because variance = npq = (mean) \( \times q \), and q must be between 0 and 1. If Mean = np = 6 and Variance = npq = 9, then q = 9/6 = 1.5, which is impossible since probability cannot exceed 1. This violates the fundamental requirement that 0 ≤ q ≤ 1.
In simple words: In any binomial distribution, the variance can never be bigger than the mean. This statement tries to claim variance is larger than mean, which breaks the rules.

Exam Tip: Always verify that q (the failure probability) lies between 0 and 1 - this is an instant check for validity of a proposed binomial distribution.

Question 1. Mark (√) against the correct answer in each of the following: If A and B are mutually exclusive events such that P(A) = 0.4, P(B) = x and P(A ∪ B) = 0.5, then x = ?
(A) 0.2
(B) 0.1
(C) \( \frac{4}{5} \)
(D) None of these
Answer: (B) 0.1
In simple words: When two events cannot happen at the same time (mutually exclusive), their union probability equals the sum of individual probabilities. So 0.5 = 0.4 + x gives x = 0.1.

Exam Tip: For mutually exclusive events, always use P(A ∪ B) = P(A) + P(B) without subtracting the intersection.

Question 2. Mark (√) against the correct answer in each of the following: If A and B are independent events such that P(A) = 0.4, P(B) = x and P(A ∪ B) = 0.5, then x = ?
(A) \( \frac{4}{5} \)
(B) 0.1
(C) \( \frac{1}{6} \)
(D) None of these
Answer: (C) \( \frac{1}{6} \)
In simple words: When events are independent, their intersection is the product of individual probabilities. Using P(A ∪ B) = P(A) + P(B) - P(A)P(B), we solve 0.5 = 0.4 + x - 0.4x to get x = 1/6.

Exam Tip: For independent events, always include the -P(A ∩ B) term in the union formula, where P(A ∩ B) = P(A)P(B).

Question 3. Mark (√) against the correct answer in each of the following: If P(A) = 0.8, P(B) = 0.5 and P(B/A) = 0.4, then P(A/B) = ?
(A) 0.32
(B) 0.64
(C) 0.16
(D) 0.25
Answer: (B) 0.64
In simple words: First find P(A ∩ B) using P(B/A) = P(A ∩ B) / P(A), which gives 0.4 = P(A ∩ B) / 0.8, so P(A ∩ B) = 0.32. Then use P(A/B) = P(A ∩ B) / P(B) = 0.32 / 0.5 = 0.64.

Exam Tip: The intersection probability P(A ∩ B) is the key link between different conditional probabilities - always find it first.

Question 4. Mark (√) against the correct answer in each of the following: If P(A) = \( \frac{6}{11} \), P(B) = \( \frac{5}{11} \) and P(A ∪ B) = \( \frac{7}{11} \), then P(A/B) = ?
(A) \( \frac{5}{6} \)
(B) \( \frac{5}{7} \)
(C) \( \frac{6}{7} \)
(D) \( \frac{4}{5} \)
Answer: (D) \( \frac{4}{5} \)
In simple words: From the union formula, P(A ∪ B) = P(A) + P(B) - P(A ∩ B), we get 7/11 = 6/11 + 5/11 - P(A ∩ B), so P(A ∩ B) = 4/11. Then P(A/B) = (4/11) / (5/11) = 4/5.

Exam Tip: Always rearrange the union formula to isolate P(A ∩ B) before computing conditional probabilities.

Question 5. Mark (√) against the correct answer in each of the following: If A and B are events such that P(A) = \( \frac{1}{2} \), P(B) = \( \frac{7}{12} \) and P(A' ∪ B') = \( \frac{1}{4} \), then A and B are
(A) Independent
(B) Mutually exclusive
(C) Both 'a' and 'b'
(D) None of these
Answer: (D) None of these
In simple words: Using P(A' ∪ B') = 1 - P(A ∩ B), we find P(A ∩ B) = 3/4. For independence, we check: P(A)P(B) = (1/2)(7/12) = 7/24, which is not 3/4, so they are not independent. For mutual exclusivity, P(A ∩ B) should be 0, which it is not. Therefore, neither property holds.

Exam Tip: Use De Morgan's Law: P(A' ∪ B') = P'(A ∩ B) = 1 - P(A ∩ B) to convert complement union problems.

Question 6. Mark (√) against the correct answer in each of the following: It is given that the probability that A can solve a given problem is \( \frac{3}{5} \) and the probability that B can solve the same problem is \( \frac{2}{3} \). The probability that at least one of A and B can solve a problem is
(A) \( \frac{2}{5} \)
(B) \( \frac{1}{15} \)
(C) \( \frac{13}{15} \)
(D) \( \frac{2}{15} \)
Answer: (C) \( \frac{13}{15} \)
In simple words: Since the events are independent, P(at least one solves) = P(A solves) + P(B solves) - P(both solve) = 3/5 + 2/3 - (3/5)(2/3) = 9/15 + 10/15 - 6/15 = 13/15.

Exam Tip: "At least one" means we use the union formula, and don't forget to subtract the product for independent events.

Question 7. Mark (√) against the correct answer in each of the following: The probabilities of A, B and C of solving a problem are \( \frac{1}{6} \), \( \frac{1}{5} \) and \( \frac{1}{3} \) respectively. What is the probability that the problem is solved?
(A) \( \frac{4}{9} \)
(B) \( \frac{5}{9} \)
(C) \( \frac{1}{3} \)
(D) None of these
Answer: (B) \( \frac{5}{9} \)
In simple words: For three independent events, it is easier to use P(at least one solves) = 1 - P(none solves) = 1 - (5/6)(4/5)(2/3) = 1 - 40/90 = 50/90 = 5/9.

Exam Tip: With three or more independent events, the complement rule is usually much faster than the full inclusion-exclusion formula.

Question 8. Mark (√) against the correct answer in each of the following: A can hit a target 4 times in 5 shots, B can hit 3 times in 4 shots, and C can hit 2 times in 3 shots. The probability that B and C hit and A does not hit is
(A) \( \frac{1}{10} \)
(B) \( \frac{2}{5} \)
(C) \( \frac{7}{12} \)
(D) None of these
Answer: (A) \( \frac{1}{10} \)
In simple words: P(A) = 4/5, P(B) = 3/4, P(C) = 2/3. We need P(B ∩ C ∩ A') = P(B)P(C)P(A') = (3/4)(2/3)(1/5) = 6/60 = 1/10.

Exam Tip: When finding the probability of a specific combination of hits and misses, multiply each individual probability for independent events.

Question 9. Mark (√) against the correct answer in each of the following: A machine operates only when all of its three components function. The probabilities of the failures of the first, second and third component are 0.2, 0.3 and 0.5, respectively. What is the probability that the machine will fail?
(A) 0.70
(B) 0.72
(C) 0.07
(D) None of these
Answer: (B) 0.72
In simple words: The machine works only if all three components work. P(all work) = (1 - 0.2)(1 - 0.3)(1 - 0.5) = (0.8)(0.7)(0.5) = 0.28. So P(machine fails) = 1 - 0.28 = 0.72.

Exam Tip: For systems that require all components to function, use the complement approach: find probability all work, then subtract from 1.

Question 10. Mark (√) against the correct answer in each of the following: A die is rolled. If the outcome is an odd number, what is the probability that it is prime?
(A) \( \frac{2}{3} \)
(B) \( \frac{3}{4} \)
(C) \( \frac{5}{12} \)
(D) None of these
Answer: (A) \( \frac{2}{3} \)
In simple words: Given the outcome is odd, the possibilities are {1, 3, 5}. Of these, {3, 5} are prime. So P(prime | odd) = 2/3.

Exam Tip: In conditional probability problems, the given condition redefines the sample space - work only within that restricted set.

Question 11. Mark (√) against the correct answer in each of the following: If A and B are events such that P(A) = 0.3, P(B) = 0.2 and P(A ∩ B) = 0.1, then \( P(\bar{A} \cap B) \) = ?
(A) 0.2
(B) 0.1
(C) 0.4
(D) 0.5
Answer: (B) 0.1
In simple words: The event \( \bar{A} \cap B \) represents "not A but B happens". So \( P(\bar{A} \cap B) \) = P(B) - P(A ∩ B) = 0.2 - 0.1 = 0.1.

Exam Tip: Remember that \( \bar{A} \cap B \) = B - (A ∩ B), so subtract the intersection from the total of B.

Question 13. Mark (√) against the correct answer in each of the following: If A and B are events such that P(A) = 0.4, P(B) = 0.8 and P(B/A) = 0.6, then P(A/B) = ?
(A) 0.2
(B) 0.3
(C) 0.4
(D) 0.5
Answer: (B) 0.3
In simple words: From P(B/A) = P(A ∩ B) / P(A) = 0.6, we find P(A ∩ B) = 0.6 × 0.4 = 0.24. Then P(A/B) = P(A ∩ B) / P(B) = 0.24 / 0.8 = 0.3.

Exam Tip: Always isolate P(A ∩ B) first from the given conditional probability before finding the reverse conditional.

Question 14. Mark (√) against the correct answer in each of the following: If A and B are independent events, then \( P(\bar{A}/\bar{B}) \) = ?
(A) 1 - P(A)
(B) 1 - P(B)
(C) 1 - P(A/\bar{B})
(D) 1 - P(\bar{A}/B)
Answer: (A) 1 - P(A)
In simple words: For independent events, \( P(\bar{A}/\bar{B}) = \frac{P(\bar{A} \cap \bar{B})}{P(\bar{B})} = \frac{P(\bar{A})P(\bar{B})}{P(\bar{B})} = P(\bar{A}) = 1 - P(A) \). Independence is preserved under complementation.

Exam Tip: If A and B are independent, then A and \(\bar{B}\), \(\bar{A}\) and B, and \(\bar{A}\) and \(\bar{B}\) are all independent pairs.

Question 15. Mark (√) against the correct answer in each of the following: If A and B are two events such that P(A ∪ B) = \( \frac{5}{6} \), P(A ∩ B) = \( \frac{1}{3} \) and P(\bar{B}) = \( \frac{1}{2} \), then the events A and B are
(A) Independent
(B) Dependent
(C) Mutually exclusive
(D) None of these
Answer: (A) Independent
In simple words: From P(\bar{B}) = 1/2, we get P(B) = 1/2. Using the union formula, P(A) = P(A ∪ B) - P(B) + P(A ∩ B) = 5/6 - 1/2 + 1/3 = 2/3. Checking independence: P(A)P(B) = (2/3)(1/2) = 1/3 = P(A ∩ B). Since the product of individual probabilities equals the intersection probability, A and B are independent.

Exam Tip: Always verify independence by checking if P(A ∩ B) = P(A)P(B) - this is the defining condition.

Question 16. Mark (√) against the correct answer in each of the following: A die is thrown twice, and the sum of the numbers appearing is observed to be 7. What is the conditional probability that the number 2 has appeared at least once?
(A) \( \frac{1}{6} \)
(B) \( \frac{1}{3} \)
(C) \( \frac{2}{7} \)
(D) \( \frac{3}{5} \)
Answer: (B) \( \frac{1}{3} \)
In simple words: The pairs that sum to 7 are {(1,6), (2,5), (3,4), (4,3), (5,2), (6,1)}. Among these six outcomes, the number 2 appears in exactly two pairs: (2,5) and (5,2). Therefore, P(2 appears | sum = 7) = 2/6 = 1/3.

Exam Tip: In conditional probability with a given event, restrict your sample space to only those outcomes satisfying the condition.

Question 17. Mark (√) against the correct answer in each of the following: Two numbers are selected at random from integers 1 through 9. If the sum is even, what is the probability that both numbers are odd?
(A) \( \frac{1}{6} \)
(B) \( \frac{2}{3} \)
(C) \( \frac{4}{9} \)
(D) \( \frac{5}{8} \)
Answer: (D) \( \frac{5}{8} \)
In simple words: A sum is even only when both numbers are even or both are odd. From 1 to 9, there are 5 odd numbers and 4 even numbers. Cases with even sum: both odd = \( ^5C_1 \times ^4C_1 = 20 \) and both even = \( ^4C_1 \times ^3C_1 = 12 \), total = 32. Favorable (both odd) = 20. So P(both odd | even sum) = 20/32 = 5/8.

Exam Tip: For conditional probability with selection problems, first count all outcomes matching the condition, then count favorable outcomes within that set.

Question 18. Mark (√) against the correct answer in each of the following: In a class, 60% of the students read mathematics, 25% biology and 15% both mathematics and biology. One student is selected at random. What is the probability that he reads mathematics if it is known that he reads biology?
(A) \( \frac{2}{5} \)
(B) \( \frac{3}{5} \)
(C) \( \frac{3}{8} \)
(D) \( \frac{5}{8} \)
Answer: (B) \( \frac{3}{5} \)
In simple words: Given: P(Math) = 0.6, P(Bio) = 0.25, P(Math ∩ Bio) = 0.15. We need P(Math | Bio) = P(Math ∩ Bio) / P(Bio) = 0.15 / 0.25 = 15/25 = 3/5.

Exam Tip: When given a percentage overlap (students studying both), this directly gives you the intersection probability.

Question 19. Mark (√) against the correct answer in each of the following: A couple has 2 children. What is the probability that both are boys, if it is known that one of them is a boy?
(A) \( \frac{1}{3} \)
(B) \( \frac{2}{3} \)
(C) \( \frac{3}{4} \)
(D) \( \frac{1}{4} \)
Answer: (A) \( \frac{1}{3} \)
In simple words: Given that at least one child is a boy, the possible outcomes are: (B, G), (G, B), and (B, B). Of these three equally likely scenarios, only one has both children as boys. Therefore, P(both boys | at least one boy) = 1/3.

Exam Tip: When a condition reduces the sample space, enumerate all remaining possibilities and count directly.

Question 20. Mark (√) against the correct answer in each of the following: An unbiased die is tossed twice. What is the probability of getting a 4, 5 or 6 on the first toss and a 1, 2, 3 or 4 on the second toss?
(A) \( \frac{1}{3} \)
(B) \( \frac{2}{3} \)
(C) \( \frac{3}{4} \)
(D) \( \frac{5}{6} \)
Answer: (A) \( \frac{1}{3} \)
In simple words: On the first toss, P(4, 5, or 6) = 3/6 = 1/2. On the second toss, P(1, 2, 3, or 4) = 4/6 = 2/3. Since the tosses are independent, multiply: (1/2) × (2/3) = 1/3.

Exam Tip: For independent sequential events, multiply the individual probabilities together.

Question 21. Mark (√) against the correct answer in each of the following: A fair coin is tossed 6 times. What is the probability of getting at least 3 heads?
(A) \( \frac{11}{16} \)
(B) \( \frac{21}{32} \)
(C) \( \frac{1}{18} \)
(D) \( \frac{3}{64} \)
Answer: (B) \( \frac{21}{32} \)
In simple words: Using the binomial distribution with n = 6, p = 1/2, q = 1/2, we calculate P(X ≥ 3) = P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6). Each term uses the formula \( ^nC_x p^x q^{n-x} \). After computing all four terms, the total is 21/32.

Exam Tip: For binomial problems, organize your calculation by setting up each term clearly and then summing them systematically.

Question 22. Mark (√) against the correct answer in each of the following: A coin is tossed 5 times. What is the probability that tail appears an odd number of times?
(A) \( \frac{3}{5} \)
(B) \( \frac{2}{15} \)
(C) \( \frac{1}{2} \)
(D) \( \frac{1}{3} \)
Answer: (C) \( \frac{1}{2} \)
In simple words: We need P(1 tail) + P(3 tails) + P(5 tails). With n = 5 and p = q = 1/2, each term becomes \( ^5C_x (1/2)^5 \). Computing \( ^5C_1 + ^5C_3 + ^5C_5 = 5 + 10 + 1 = 16 \), giving probability 16/32 = 1/2.

Exam Tip: For fair coins with p = q = 1/2, the total of \( ^nC_k \) for all favorable k values divided by \( 2^n \) gives the answer quickly.

Question 23. Mark (√) against the correct answer in each of the following: A coin is tossed 5 times. What is the probability that the head appears an even number of times?
(A) \( \frac{2}{5} \)
(B) \( \frac{3}{5} \)
(C) \( \frac{4}{15} \)
(D) \( \frac{1}{2} \)
Answer: (D) \( \frac{1}{2} \)
In simple words: An even number of heads means 0, 2, or 4 heads. Using the binomial formula with n = 5 and p = q = 1/2, we compute P(X = 0) + P(X = 2) + P(X = 4) = \( (^5C_0 + ^5C_2 + ^5C_4) / 2^5 = (1 + 10 + 5) / 32 = 16/32 = 1/2 \).

Exam Tip: Due to symmetry in fair coin problems, P(even heads) = P(odd heads) = 1/2 always.

Question 24. Mark (√) against the correct answer in each of the following: 8 coins are tossed simultaneously. The probability of getting at least 6 heads is
(a) \( \frac{7}{64} \)
(b) \( \frac{57}{64} \)
(c) \( \frac{37}{256} \)
(d) \( \frac{249}{256} \)
Answer: (c) \( \frac{37}{256} \)
To solve this, apply Bernoulli's Trial formula: \( P(\text{Success}=x) = nCx \cdot p^x \cdot q^{(n-x)} \) where \( x = 0, 1, 2, \ldots, n \) and \( q = (1-p) \). Since the coin is tossed 8 times, the total number of outcomes is \( 2^8 \). The favorable outcomes occur when getting at least 6 heads means getting 6, 7, or 8 heads. The probability of success (getting a head) is \( \frac{1}{2} \) and the probability of failure is also \( \frac{1}{2} \). Therefore, the probability of getting at least 6 heads is computed as:
\( P(6) + P(7) + P(8) = 8C6 \left(\frac{1}{2}\right)^6 \left(\frac{1}{2}\right)^2 + 8C7 \left(\frac{1}{2}\right)^7 \left(\frac{1}{2}\right)^1 + 8C8 \left(\frac{1}{2}\right)^8 \left(\frac{1}{2}\right)^0 = \frac{28 + 8 + 1}{256} = \frac{37}{256} \)
In simple words: When you flip 8 coins, find how many ways you can get 6, 7, or 8 heads. Add these possibilities and divide by the total number of outcomes (256) to get your answer.

Exam Tip: Always identify whether the question asks for "at least" (meaning ≥) or "at most" (meaning ≤), then sum the correct probabilities accordingly.

Question 25. Mark (√) against the correct answer in each of the following: A die is thrown 5 times. If getting an odd number is a success, then what is the probability of getting at least 4 successes?
(a) \( \frac{4}{5} \)
(b) \( \frac{7}{16} \)
(c) \( \frac{3}{16} \)
(d) \( \frac{3}{20} \)
Answer: (c) \( \frac{3}{16} \)
Apply Bernoulli's Trial formula: \( P(\text{Success}=x) = nCx \cdot p^x \cdot q^{(n-x)} \) where \( x = 0, 1, 2, \ldots, n \) and \( q = (1-p) \). When the die is thrown 5 times, the total number of outcomes is \( 6^5 \). The favorable outcomes for getting at least 4 successes occur when rolling 1, 3, or 5 (odd numbers). Each odd number has a probability of \( \frac{1}{6} \), so the total probability is \( p = \frac{3}{6} = \frac{1}{2} \) and \( q = \frac{1}{2} \). The probability of success is \( \frac{1}{2} \) and the probability of failure is also \( \frac{1}{2} \). Therefore, the probability of getting at least 4 successes is:
\( P(4) + P(5) = 5C4 \left(\frac{1}{2}\right)^4 \left(\frac{1}{2}\right)^1 + 5C5 \left(\frac{1}{2}\right)^5 \left(\frac{1}{2}\right)^0 = \frac{5 + 1}{32} = \frac{3}{16} \)
In simple words: Throwing a die 5 times and getting an odd number (1, 3, or 5) is your success. Find the chance of getting 4 or 5 odd numbers, then add those probabilities together.

Exam Tip: Recognize that for a standard die, odd numbers comprise exactly half the possible outcomes, making p = 1/2 and simplifying your calculations significantly.

Question 26. Mark (√) against the correct answer in each of the following: In 4 throws of a pair of dice, what is the probability of throwing doublets at least twice?
(a) \( \frac{7}{36} \)
(b) \( \frac{17}{144} \)
(c) \( \frac{19}{144} \)
(d) None of these
Answer: (c) \( \frac{19}{144} \)
Using Bernoulli's Trial formula: \( P(\text{Success}=x) = nCx \cdot p^x \cdot q^{(n-x)} \) where \( x = 0, 1, 2, \ldots, n \) and \( q = (1-p) \). For obtaining at least doublets twice when throwing a pair of dice 4 times, successes occur when rolling matching numbers. The favorable outcomes for a doublet are 6 pairs: (1,1), (2,2), (3,3), (4,4), (5,5), (6,6). Out of 36 possible outcomes, \( p = \frac{6}{36} = \frac{1}{6} \) and \( q = \frac{5}{6} \). The probability of success is \( \frac{1}{6} \) and the probability of failure is also \( \frac{5}{6} \). Therefore, the probability of getting at least 2 successes is:
\( P(2) + P(3) + P(4) = 4C2 \left(\frac{1}{6}\right)^2 \left(\frac{5}{6}\right)^2 + 4C3 \left(\frac{1}{6}\right)^3 \left(\frac{5}{6}\right)^1 + 4C4 \left(\frac{1}{6}\right)^4 \left(\frac{5}{6}\right)^0 = \frac{19}{144} \)
In simple words: A doublet means both dice show the same number. Out of 36 ways to roll two dice, only 6 are doublets. Calculate the chance of getting 2, 3, or 4 doublets in 4 throws and add them up.

Exam Tip: When calculating "at least," remember to include all cases from the minimum threshold up to the maximum possible, being careful not to omit any intermediate values.

Question 27. Mark (√) against the correct answer in each of the following: A pair of dice is thrown 7 times. If getting a total of 7 is considered a success, what is the probability of getting at most 6 successes?
(a) \( \left(\frac{5}{7}\right)^7 \)
(b) \( \left(\frac{1}{6}\right)^7 \)
(c) \( \left(1 - \frac{1}{6}\right)^7 \)
(d) None of these
Answer: (c) \( \left(1 - \frac{1}{6}\right)^7 \)
Using Bernoulli's Trial formula: \( P(\text{Success}=x) = nCx \cdot p^x \cdot q^{(n-x)} \) where \( x = 0, 1, 2, \ldots, n \) and \( q = (1-p) \), here \( n = 7 \). The favorable outcomes for getting at most 6 successes occur when rolling a total of 7. You can get a total of 7 from the pairs: (1,6), (2,5), (3,4), (4,3), (5,2), (6,1). So \( p = \frac{6}{36} = \frac{1}{6} \) and \( q = \frac{5}{6} \). The probability of success is \( \frac{1}{6} \) and the probability of failure is also \( \frac{5}{6} \). Therefore, the probability of getting at most 6 successes equals:
\( P(0) + P(1) + P(2) + P(3) + P(4) + P(5) + P(6) = 1 - P(7) = 1 - 7C7 \left(\frac{1}{6}\right)^7 \left(\frac{5}{6}\right)^0 = 1 - \left(\frac{1}{6}\right)^7 \)
In simple words: Getting a sum of 7 with two dice happens 6 ways out of 36. Instead of adding 7 separate probabilities, it is easier to subtract the chance of getting all 7 successes from 1.

Exam Tip: For "at most" problems with large n, use the complement rule - compute 1 minus the probability of exceeding your threshold, which is much faster than summing many terms.

Question 28. Mark (√) against the correct answer in each of the following: The probability that a man can hit a target is \( \frac{3}{4} \). He tries five times. What is the probability that he will hit the target at least 3 times?
(a) \( \frac{459}{512} \)
(b) \( \frac{291}{364} \)
(c) \( \frac{371}{464} \)
(d) None of these
Answer: (a) \( \frac{459}{512} \)
The probability that the man hits the target is \( \frac{3}{4} \). Using Bernoulli's Trial, we have \( P(\text{Success}=x) = nCx \cdot p^x \cdot q^{(n-x)} \) where \( x = 0, 1, 2, \ldots, n \) and \( q = (1-p) \), with \( n = 5 \). Here \( p = \frac{3}{4} \) and \( q = \frac{1}{4} \). The probability that he will hit at least 3 times is:
\( P(3) + P(4) + P(5) = 5C3 \left(\frac{3}{4}\right)^3 \left(\frac{1}{4}\right)^2 + 5C4 \left(\frac{3}{4}\right)^4 \left(\frac{1}{4}\right)^1 + 5C5 \left(\frac{3}{4}\right)^5 \left(\frac{1}{4}\right)^0 = \frac{459}{512} \)
In simple words: The man has a 3 in 4 chance of hitting the target each time. Over 5 attempts, calculate how likely he is to succeed 3, 4, or 5 times, then add those chances together.

Exam Tip: When probabilities are not 1/2, compute each term carefully using the binomial coefficients and exact powers, as rounding errors can lead to wrong answer choices.

Question 29. Mark (√) against the correct answer in each of the following: The probability of the safe arrival of one ship out of 5 is \( \frac{1}{5} \). What is the probability of the safe arrival of at least 3 ships?
(a) \( \frac{1}{31} \)
(b) \( \frac{3}{52} \)
(c) \( \frac{181}{3125} \)
(d) \( \frac{184}{3125} \)
Answer: (c) \( \frac{181}{3125} \)
The probability of safe arrival of the ship is \( \frac{1}{5} \). Using Bernoulli's Trial, we have \( P(\text{Success}=x) = nCx \cdot p^x \cdot q^{(n-x)} \) where \( x = 0, 1, 2, \ldots, n \) and \( q = (1-p) \), with \( n = 5 \). Here \( p = \frac{1}{5} \) and \( q = \frac{4}{5} \). The probability of safe arrival of at least 3 ships is:
\( P(3) + P(4) + P(5) = 5C3 \left(\frac{1}{5}\right)^3 \left(\frac{4}{5}\right)^2 + 5C4 \left(\frac{1}{5}\right)^4 \left(\frac{4}{5}\right)^1 + 5C5 \left(\frac{1}{5}\right)^5 \left(\frac{4}{5}\right)^0 = \frac{181}{3125} \)
In simple words: Each ship has a 1 in 5 chance of arriving safely. Find the likelihood that 3, 4, or 5 ships (out of 5 total) will arrive safely, then sum those probabilities.

Exam Tip: Notice that when p is small (like 1/5 here), the probability of "at least 3" successes in only 5 trials tends to be quite small - verify your arithmetic to avoid choosing an unreasonably large answer.

Question 30. Mark (√) against the correct answer in each of the following: The probability that an event E occurs in one trial is 0.4. Three independent trials of the experiment are performed. What is the probability that E occurs at least once?
(a) 0.784
(b) 0.936
(c) 0.964
(d) None of these
Answer: (a) 0.784
The probability of occurrence of an event E in one trial is 0.4. Using Bernoulli's Trial, we have \( P(\text{Success}=x) = nCx \cdot p^x \cdot q^{(n-x)} \) where \( x = 0, 1, 2, \ldots, n \) and \( q = (1-p) \), with \( n = 3 \). Here \( p = 0.4 \) and \( q = 0.6 \). The probability that E occurs at least once is:
\( P(1) + P(2) + P(3) = 3C1 (0.4)^1 (0.6)^2 + 3C2 (0.4)^2 (0.6)^1 + 3C3 (0.4)^3 (0.6)^0 = \frac{98}{125} = 0.784 \)
In simple words: An event has a 40% chance of happening in each trial. Over 3 independent tries, find the probability it happens at least once by adding up the chances of it occurring exactly 1, 2, or 3 times.

Exam Tip: For "at least once" problems, the complement shortcut (1 - P(0)) is often faster than summing multiple terms, especially when dealing with decimal probabilities.