CBSE Class 10 Maths HOTs Probability Set 05

Question. Complete the following statements:
(i) Probability of an event E + Probability of the event ‘not E’ = ___________.
(ii) The probability of an event that cannot happen is ___________. Such an event is called ___________.
(iii) The probability of an event that is certain to happen is ___________. Such an event is called ___________.
(iv) The sum of the probabilities of all the elementary events of an experiment is ___________.
(v) The probability of an event is greater than or equal to ___________ and less than or equal to ___________.

Answer:
(i) 1.
(ii) 0. Such an event is called impossible event.
(iii) 1. Such an event is called sure event.
(iv) 1.
(v) 0 and 1.

Question. Which of the following experiments have equally likely outcomes? Explain.
(i) A driver attempts to start a car. The car starts or does not start.
(ii) A player attempts to shoot a basketball. She/he shoots or misses the shot.
(iii) A trial is made to answer a true-false question. The answer is right or wrong.
(iv) A baby is born. It is a boy or a girl.

Answer:
(i) The outcome is not equally likely because the car starts normally only when there is some defect, the car does not start.
(ii) The outcome is not equally likely because the outcome depends on the training of the player.
(iii) The outcome in the trial of true-false question is, either true or false. Hence, the two outcomes are equally likely.
(iv) A baby can be either a boy or a girl and both the outcomes have equally likely chances.
So, experiments (iii) and (iv) have equally likely outcomes.

Question. Why is tossing a coin considered to be a fair way of deciding which team should get the ball at the beginning of a football game?

Answer: When we toss a coin, the outcomes head and tail are equally likely. So, the result of an individual coin toss is completely unpredictable.

Question. Which of the following cannot be the probability of an event?
(A) \( \frac{2}{3} \)
(B) – 1.5
(C) 15%
(D) 0.7

Answer: (B) – 1.5

Question. If P (E) = 0.05, what is the probability of ‘not E’?

Answer: We have P (E) + P (not E) = 1
P (E) = 0.05
P (not E) = 1 – 0.05 = 0.95

Question. A bag contains lemon flavoured candies only. Malini takes out one candy without looking into the bag. What is the probability that she takes out
(i) an orange flavoured candy?
(ii) a lemon flavoured candy?

Answer:
(i) Because a bag contains only lemon flavoured candies
\( \implies \) P (an orange flavoured candy) = 0
(ii) P (a lemon flavoured candy) = 1

Question. It is given that in a group of 3 students, the probability of 2 students not having the same birthday is 0.992. What is the probability that the 2 students have the same birthday?

Answer: We have P (E) + P (not E) = 1
\( \implies \) P (E) + 0.992 = 1
\( \implies \) P (E) = 1 – 0.992 = 0.008

Question. A bag contains 3 red balls and 5 black balls. A ball is drawn at random from the bag. What is the probability that the ball drawn is (i) red? (ii) not red?

Answer: Number of red balls = 3
Number of black balls = 5
Total number of balls = 3 + 5 = 8
(i) P (red ball) = \( \frac{Number\ of\ red\ balls}{Total\ number\ of\ balls} = \frac{3}{8} \)
(ii) P (not red) = 1 – \( \frac{3}{8} = \frac{5}{8} \)

Question. A box contains 5 red marbles, 8 white marbles and 4 green marbles. One marble is taken out of the box at random. What is the probability that the marble taken out will be (i) red? (ii) white? (iii) not green?

Answer: Total number of marbles = 5 + 8 + 4 = 17
(i) P (red marble) = \( \frac{5}{17} \)
(ii) P (white marble) = \( \frac{8}{17} \)
(iii) P (not a green marble) = 1 – \( \frac{4}{17} = \frac{13}{17} \)

Question. A piggy bank contains hundred 50p coins, fifty Rs. 1 coins, twenty Rs. 2 coins and ten Rs. 5 coins. If it is equally likely that one of the coins will fall out when the bank is turned upside down, what is the probability that the coin
(i) will be a 50 p coin?
(ii) will not be a Rs. 5 coin?

Answer: Number of 50 p coins = 100
Number of Rs. 1 coins = 50
Number of Rs. 2 coins = 20
Number of Rs. 5 coins = 10
Total number of coins = 180
(i) P (50 p coin) = \( \frac{100}{180} = \frac{5}{9} \)
(ii) P (not a Rs. 5 coin) = 1 – \( \frac{10}{180} = \frac{170}{180} = \frac{17}{18} \)

Question. A die is thrown once. Find the probability of getting
(i) a prime number.
(ii) a number lying between 2 and 6.
(iii) an odd number.

Answer: (i) Prime numbers on a die = 2, 3, 5
\( \therefore \) P (a prime number) = \( \frac{3}{6} = \frac{1}{2} \)
(ii) Number lying between 2 and 6 = 3, 4, 5
\( \therefore \) P (a number lying between 2 and 6) = \( \frac{3}{6} = \frac{1}{2} \)
(iii) Odd numbers = 1, 3, 5
\( \therefore \) P (an odd number) = \( \frac{3}{6} = \frac{1}{2} \)

Question. One card is drawn from a well-shuffled deck of 52 cards. Find the probability of getting
(i) a king of red colour.
(ii) a face card.
(iii) a red face card.
(iv) the jack of hearts.
(v) a spade.
(vi) the queen of diamonds.

Answer: Number of cards in a well-shuffled deck = 52.
(i) P (a king of red colour) = \( \frac{2}{52} = \frac{1}{26} \)
(ii) P (a face card) = \( \frac{12}{52} = \frac{3}{13} \)
(iii) P (a red face card) = \( \frac{6}{52} = \frac{3}{26} \)
(iv) P (the jack of hearts) = \( \frac{1}{52} \)
(v) P (a spade) = \( \frac{13}{52} = \frac{1}{4} \)
(vi) P (the queen of diamonds) = \( \frac{1}{52} \)

Question. Five cards - the ten, jack, queen, king and ace of diamonds, are well shuffled with their face downwards. One card is then picked up at random.
(i) What is the probability that the card is the queen?
(ii) If the queen is drawn and put aside, what is the probability that the second card picked up is (a) an ace? (b) a queen?

Answer: Out of 5 cards there is only one queen.
(i) P (getting queen) = \( \frac{1}{5} \)
(ii) (a) P (an ace) = \( \frac{1}{4} \) [when queen is drawn, four cards are left].
(b) P (a queen) = \( \frac{0}{4} = 0 \)

Question. 12 defective pens are accidentally mixed with 132 good ones. It is not possible to just look at a pen and tell whether or not it is defective. One pen is taken out at random from this lot. Determine the probability that the pen taken out is a good one.

Answer: Number of defective pens = 12
Number of good pens = 132
Total number of pens = 12 + 132 = 144
\( \therefore \) P (the pen is good one) = \( \frac{132}{144} = \frac{11}{12} \)

Question. (i) A lot of 20 bulbs contain 4 defective ones. One bulb is drawn at random from the lot. What is the probability that this bulb is defective?
(ii) Suppose the bulb drawn in (i) is not defective and is not replaced. Now one bulb is drawn at random from the rest. What is the probability that this bulb is not defective?

Answer: (i) Total number of bulbs = 20
Number of defective bulbs = 4
\( \therefore \) (P bulb drawn is defective) = \( \frac{4}{20} = \frac{1}{5} \)
(ii) Remaining bulbs = 19
\( \therefore \) (P bulb drawn is not defective) = \( \frac{15}{19} \)

Question. A box contains 90 discs which are numbered from 1 to 90. If one disc is drawn at random from the box, find the probability that it bears
(i) a two digit number.
(ii) a perfect square number.
(iii) a number divisible by 5.

Answer: Total numbers of discs = 90
(i) P (a two digit number) = \( \frac{81}{90} = \frac{9}{10} \)
(ii) Here, Perfect square numbers are – 1, 4, 9, 16, 25, 36, 49, 64, 81
\( \therefore \) P (getting a perfect square number) = \( \frac{9}{90} = \frac{1}{10} \)
(iii) Numbers divisible by 5 are – 5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, 60, 65, 70, 75, 80, 85, 90
\( \therefore \) P (getting) a number divisible by 5 = \( \frac{18}{90} = \frac{1}{5} \)

Question. A child has a die whose six faces show the letters as given below:
A B C D E A
The die is thrown once. What is the probability of getting (i) A? (ii) D?

Answer: (i) P (getting A) = \( \frac{2}{6} = \frac{1}{3} \)
(ii) P (getting D) = \( \frac{1}{6} \)

Question. A lot consists of 144 ball pens of which 20 are defective and the others are good. Nuri will buy a pen if it is good, but will not buy if it is defective. The shopkeeper draws one pen at random and gives it to her. What is the probability that
(i) she will buy it?
(ii) she will not buy it?

Answer: Total number of ball pens = 144
Number of defective pens = 20
Number of good pens = 144 – 20 = 124
(i) P (E) = \( \frac{124}{144} = \frac{31}{36} \)
(ii) P (not E) = 1 – \( \frac{31}{36} = \frac{36 - 31}{36} = \frac{5}{36} \)

Question. Two dice one blue and one grey, are thrown at the same time. Now
(i) Complete the following table:
Event: (Sum of 2 dice) | Probability
2 | 1/36
3 |
4 |
5 |
6 |
7 |
8 | 5/36
9 |
10 |
11 |
12 | 1/36
(ii) A student argues that–there are 11 possible outcomes (2, 3, 4, 5, 6, 7, 8, 9, 10, 11 and 12). Therefore, each of them has a probability \( \frac{1}{11} \). Do you agree with this argument? Justify your answer.

Answer: (i) Total number of possible outcomes = 36
for P(sum 2) = \( \frac{1}{36} \)
(1, 2) and (2, 1) favour the event of getting the sum 3. So, P (sum 3) = \( \frac{2}{36} \)
(1,3), (2,2), (3,1) getting the sum 4. So, P (sum 4) = \( \frac{3}{36} = \frac{1}{12} \)
(1, 4), (2,3), (3,2), (4,1) favour the sum 5. So, P (sum 5) = \( \frac{4}{36} = \frac{1}{9} \)
(1,5), (2,4), (3,3), (4,2), (5,1) favour the sum 6. So, P (sum 6) = \( \frac{5}{36} \)
(1,6), (2,5), (3,4), (4,3), (5,2), (6,1) favour the sum 7. So, P (sum 7) = \( \frac{6}{36} = \frac{1}{6} \)
(3,6), (4,5), (5,4), (6,3) favour the sum 9. So, P (sum 9) = \( \frac{4}{36} = \frac{1}{9} \)
(4,6), (5,5), (6,4) favour the sum 10. So, P (sum 10) = \( \frac{3}{36} = \frac{1}{12} \)
(5,6), (6,5) favour the sum 11. So, P (sum 11) = \( \frac{2}{36} = \frac{1}{18} \)
for sum of 12, P(sum 12) = \( \frac{1}{36} \)
(ii) No, because the outcomes as 11 different sum are not equally likely.

Question. A game consists of tossing a one rupee coin 3 times and noting its outcome each time. Hanif wins if all the tosses give the same result i.e., three heads or three tails, and loses otherwise. Calculate the probability that Hanif will lose the game.

Answer: Possible outcomes are HHH, TTT, HHT, HTH, THH, TTH, THT, HTT = 8
P(win the game) = \( \frac{2}{8} = \frac{1}{4} \).
Hence P(lose the game) = 1 – \( \frac{1}{4} = \frac{3}{4} \)

Question. A die is thrown twice. What is the probability that
(i) 5 will not come up either time?
(ii) 5 will come up at least once?
[Hint: Throwing a die twice and throwing two dice simultaneously are treated as the same experiment.]

Answer: Total outcomes = 36
Number of Outcomes in favour of 5 is (1,5) (2,5) (3,5) (4,5) (5,5) (6,5) (5,1) (5,2) (5,3) (5,4) (5,6) = 11
(i) P (5 will not come up either time) = \( \frac{25}{36} \)
(ii) P (5 will come up at least once) = \( \frac{11}{36} \)

Question. Which of the following arguments are correct and which are not correct? Give reasons for your answer.
(i) If two coins are tossed simultaneously there are three possible outcomes– two heads, two tails or one of each. Therefore, for each of these outcomes, the probability is \( \frac{1}{3} \).
(ii) If a die is thrown, there are two possible outcomes– an odd number or an even number. Therefore, the probability of getting an odd number is \( \frac{1}{2} \).

Answer:
(i) Argument is incorrect. The possible outcomes are– (HH), (HT), (TH), (TT).
\( \therefore \) P (HH) = \( \frac{1}{4} \), P (TT) = \( \frac{1}{4} \), P (HT or TH) = \( \frac{2}{4} = \frac{1}{2} \)
(ii) Argument is correct. Possible outcomes = 1, 2, 3, 4, 5, 6. Odd numbers are = 1, 3, 5. Hence P (an odd number) = \( \frac{3}{6} = \frac{1}{2} \). Even numbers are = 2, 4, 6. Hence P (an even number) = \( \frac{3}{6} = \frac{1}{2} \).

Question. If an event cannot occur, then its probability is
(a) 1
(b) \( \frac{3}{4} \)
(c) \( \frac{1}{2} \)
(d) 0

Answer: (d) 0

Question. An event is very unlikely to happen. Its probability is closest to
(a) 0.0001
(b) 0.001
(c) 0.01
(d) 0.1

Answer: (a) 0.0001

Question. The probability expressed as a percentage of a particular occurrence can never be
(a) less than 100
(b) less than 0
(c) greater than 1
(d) anything but a whole number

Answer: (b) less than 0

Question. If P(A) denotes the probability of an event A, then
(a) P(A) < 0
(b) P(A) > 1
(c) 0 ≤ P(A) ≤ 1
(d) –1 ≤ P(A) ≤ 1

Answer: (c) 0 ≤ P(A) ≤ 1

Question. The probability of getting a bad egg in a lot of 400 is 0.035. The number of bad eggs in the lot is
(a) 7
(b) 14
(c) 21
(d) 28

Answer: P(a bad egg) = \( \frac{No.\ of\ bad\ eggs}{Total\ eggs} \)
\( \implies \) 0.035 = \( \frac{x}{400} \)
\( \implies \) x = 14
Answer: (b) 14

Question. A girl calculates that the probability of her winning the first prize in a lottery is 0.08. If 6000 tickets are sold, how many tickets has she bought?
(a) 40
(b) 240
(c) 480
(d) 750

Answer: 0.08 × 6000 = 480
Answer: (c) 480

Question. A child’s game has 8 triangles of which 3 are blue and rest are red, and 10 squares of which 6 are blue and rest are red. One piece is lost at random. Find the probability that it is a
(i) triangle (ii) square (iii) square of blue colour (iv) triangle of red colour

Answer: Total number of pieces = 8 + 10 = 18
(i) No. of triangles = 8. Hence, P(triangle is lost) = \( \frac{8}{18} = \frac{4}{9} \)
(ii) No. of squares = 10. Hence, P(square is lost) = \( \frac{10}{18} = \frac{5}{9} \)
(iii) No. of squares of blue colour = 6. So, P(square of blue colour is lost) = \( \frac{6}{18} = \frac{1}{3} \)
(iv) No. of triangles of red colour = 8 – 3 = 5. So, P(triangle of red colour is lost) = \( \frac{5}{18} \)

Question. In a game, the entry fee is Rs. 5. The game consists of tossing a coin 3 times. If one or two heads show, Shweta gets her entry fee back. If she throws 3 heads, she receives double the entry fees. Otherwise she will lose. For tossing a coin three times, find the probability that she
(i) loses the entry fee.
(ii) gets double entry fee.
(iii) just gets her entry fee.

Answer: Possible outcomes when a coin is tossed 3 times: HHH, HHT, HTH, THH, TTH, THT, HTT, TTT. Total no. of outcomes = 8.
(i) Shweta will lose the entry fee if she gets ‘TTT’.
\( \therefore \) P(Shweta loses the entry fee) = \( \frac{1}{8} \)
(ii) Shweta will get double the entry fee if she gets HHH.
\( \therefore \) P(Shweta will get double the entry fee) = \( \frac{1}{8} \).
(iii) Shweta will get her entry fee, if she get HHT, HTH, THH, TTH, THT or HTT. No. of ways = 6.
\( \therefore \) P(Shweta will get her entry fee) = \( \frac{6}{8} = \frac{3}{4} \).

Question. At a fete, cards bearing numbers 1 to 1000, one number on one card, are put in a box. Each player select one card at random and that card is not replaced. If the selected card has a perfect square greater than 500, the player wins a prize. What is the probability that
(i) the first player wins a prize?
(ii) the second player wins a prize, if the first has won?

Answer: Number of ways to draw a card by Ist player = 1000. Perfect square greater than 500 are 529, 576, 625, 676, 729, 784, 841, 900, 961.
\( \implies \) Probability of Ist player winning the prize = \( \frac{9}{1000} \)
When Ist has won the prize then cards left = 999 (because card is not replaced). Cards with no. which a perfect square greater than 500 = 8 (1 winning number removed).
\( \implies \) Probability of second player winning a prize when the first player has won = \( \frac{8}{999} \).

Question. There are 1000 sealed envelopes in a box, 10 of them contain a cash prize of Rs. 100 each, 100 of them contain a cash prize of Rs. 50 each and 200 of them contain a cash prize of Rs. 10 each and rest do not contain any cash prize. If they are well shuffled and an envelope is picked up out, what is the probability that it contains no cash prize?

Answer: Total number of envelopes = 1000
Let A = envelope contains no cash prize
Number of envelopes containing no cash prize = 1000 – (10 + 100 + 200) = 690
\( \therefore \) P(A) = \( \frac{690}{1000} = \frac{69}{100} = 0.69 \).

Question. A bag contains slips numbered from 1 to 100. If Fatima chooses a slip at random from the bag, it will either be an odd number or an even number. Since this situation has only two possible outcomes, so, the probability of each is \( \frac{1}{2} \). Justify.

Answer: Yes, P (even number) = \( \frac{50}{100} = \frac{1}{2} \); P (odd number) = \( \frac{50}{100} = \frac{1}{2} \).

Question. In a family having three children, there may be no girl, one girl, two girls or three girls. So, the probability of each is \( \frac{1}{4} \). Is this correct? Justify your answer.

Answer: Total cases are 8. (GGG), (GGB), (GBG), (GBB), (BGG), (BGB), (BBG), (BBB)
no girl = \( \frac{1}{8} \); one girl = \( \frac{3}{8} \); 2 girls = \( \frac{3}{8} \); 3 girls = \( \frac{1}{8} \).

Question. Apoorv throws two dice once and computes the product of the numbers appearing on the dice. Peehu throws one die and squares the number that appears on it. Who has the better chance of getting the number 36? Why?

Answer: For Apoorv, the number 36 can be obtained if he gets (6, 6).
\( \therefore \) Probability of getting (6, 6) = \( \frac{1}{36} \)
For Peehu, the number 36 can be obtained if she gets 6.
\( \therefore \) Probability of getting 6 = \( \frac{1}{6} \)
As \( \frac{1}{6} > \frac{1}{36} \) so Peehu has better chance of getting 36.

Question. A student says that if you throw a die, it will show up 1 or not 1. Therefore, the probability of getting 1 and the probability of getting ‘not 1’ each is equal to \( \frac{1}{2} \). Is this correct? Give reasons.

Answer: No, because P(1) = \( \frac{1}{6} \) but P(not 1) = P (getting 2, 3, 4, 5 or 6) = \( \frac{5}{6} \).

Question. A die has its six faces marked 0, 1, 1, 1, 6, 6. Two such dice are thrown together and the total score is recorded.
(i) How many different scores are possible?
(ii) What is the probability of getting a total of 7?

Answer: Possible outcomes:
(0, 0), (0, 1), (0, 1), (0, 1), (0, 6), (0, 6)
(1, 0), (1, 1), (1, 1), (1, 1), (1, 6), (1, 6)
(1, 0), (1, 1), (1, 1), (1, 1), (1, 6), (1, 6)
(1, 0), (1, 1), (1, 1), (1, 1), (1, 6), (1, 6)
(6, 0), (6, 1), (6, 1), (6, 1), (6, 6), (6, 6)
(6, 0), (6, 1), (6, 1), (6, 1), (6, 6), (6, 6)
(i) Different total scores are 0, 1, 6, 2, 7 or 12
(ii) Let A = getting a total of 7. No. of favourable outcomes = 12
\( \therefore \) P(A) = \( \frac{12}{36} = \frac{1}{3} \)

Question. A lot consists of 48 mobiles phones of which 42 are good, 3 have only minor defects and 3 have major defects. Varnika will buy a phone if it is good but the trader will only buy a mobile if it has no major defect. One phone is selected at random from the lot. What is the probability that it is
(i) acceptable to Varnika?
(ii) acceptable to the trader?

Answer: Total number of phones = 48
Let A = phone is good. Number of good phones = 42.
\( \therefore \) P(A) = \( \frac{42}{48} = \frac{7}{8} \)
\( \implies \) Probability that Varnika will buy a phone = \( \frac{7}{8} \)
Let B = Phone has no major defect. Number of phones having no major defects = 48 – 3 = 45
P(B) = \( \frac{45}{48} = \frac{15}{16} \)
\( \implies \) Probability that phone is acceptable to the trader = \( \frac{15}{16} \)