Selina Concise Solutions for ICSE Class 7 Mathematics Chapter 22 Probability

Exercise 22(A)

Question 1. A coin is tossed once. Find the probability of
(i) getting a head
(ii) not getting a head
Answer:
When we flip a coin, the possible results are either Head (H) or Tail (T). This gives us 2 total outcomes.
(i) Probability of obtaining a head:
\( P(\text{Getting a head}) = \frac{1}{2} \)
(ii) Probability of not obtaining a head:
\( P(\text{Not getting a head}) = \frac{1}{2} \)
In simple words: A coin has two sides, so any flip has a 1 in 2 chance of landing on heads, and a 1 in 2 chance of not landing on heads.

Exam Tip: Always state the list of all possible outcomes clearly to show the examiner your base of calculation.

Question 2. A coin is tossed 80 times and the head is obtained 38 times. Now, if a coin tossed once, what will the probability of getting:
(i) a tail
(ii) ahead
Answer:
We know that the coin is flipped 80 times in total.
The number of times a head comes up is 38.
So, the number of times a tail comes up is \( 80 - 38 = 42 \).
(i) Probability of obtaining a tail:
\( P(\text{getting a tail}) = \frac{42}{80} = \frac{21}{40} \)
(ii) Probability of obtaining a head:
\( P(\text{getting a head}) = \frac{38}{80} = \frac{19}{40} \)
In simple words: To find each probability, divide the count of heads or tails by the total flips of 80, and then reduce the fraction.

Exam Tip: Always check if your fractions can be reduced to simpler terms to get full marks.

Question 3. A dice is thrown 20 times and the outcomes are noted as shown below :

Now a dice is thrown at random, find the probability of getting :
(i) 3
(ii) a number less than 3
(iii) a number greater than 3
Answer:
The total number of outcomes from throwing the die is 20.
(i) The outcome of getting 3 occurs 4 times.
\( P(\text{getting 3}) = \frac{4}{20} = \frac{1}{5} \)
(ii) The outcomes smaller than 3 are 1 and 2.
These values occur \( 2 + 3 = 5 \) times.
\( P(\text{getting a number less than 3}) = \frac{5}{20} = \frac{1}{4} \)
(iii) The outcomes larger than 3 are 4, 5, and 6.
These values occur \( 4 + 3 + 4 = 11 \) times.
\( P(\text{getting a number greater than 3}) = \frac{11}{20} \)
In simple words: Count how many times your desired numbers show up in the table. Divide that by 20 and simplify the fraction.

Exam Tip: For composite outcomes like "less than 3", first list the favorable numbers (1 and 2) before adding their frequencies.

Question 4. A survey of 50 boys showed that 21 like tea while 29 dislike it. Out of these boys, one boy is chosen at random. What is the probability that the chosen boy
(i) likes tea
(ii) dislikes tea
Answer:
The total group of boys in the survey is 50.
The number of boys who enjoy tea is 21.
The number of boys who do not enjoy tea is 29.
(i) Probability that the chosen boy likes tea:
\( P(\text{boy likes tea}) = \frac{21}{50} \)
(ii) Probability that the chosen boy dislikes tea:
\( P(\text{boy dislikes tea}) = \frac{29}{50} \)
In simple words: Put the number of boys who like or dislike tea over 50, which is the total count of boys.

Exam Tip: Remember that the probabilities of all possible options in a scenario always add up to exactly 1.

Question 5. In a cricket match, a batsman hits a boundary 12 times out of 80 balls he plays, further, if he plays one ball more, what will be the probability that:
(i) he hits a boundary
(ii) he does not hit a boundary
Answer:
The total count of balls faced by the batsman is 80.
The count of balls on which he hit a boundary is 12.
(i) Probability that he hits a boundary:
\( P(\text{Hitting a boundary}) = \frac{12}{80} = \frac{3}{20} \)
(ii) The count of balls where he did not hit a boundary is \( 80 - 12 = 68 \).
Probability that he does not hit a boundary:
\( P(\text{Not hitting a boundary}) = \frac{68}{80} = \frac{17}{20} \)
In simple words: The batsman scored boundaries on 12 balls and missed on 68 balls. We divide these numbers by the 80 total balls.

Exam Tip: You can also find the probability of the complementary event by subtracting the first probability from 1.

Question 6. There are 8 marbles in a bag with numbers from 1 to 8 marked on each of them. What is the probability of drawing a marble with number
(i) 3
(ii) 7
Answer:
The total number of marbles inside the bag is 8.
(i) Since there is only one marble with the number 3:
\( P(\text{getting a marble with number 3}) = \frac{1}{8} \)
(ii) Since there is only one marble with the number 7:
\( P(\text{getting a marble with number 7}) = \frac{1}{8} \)
In simple words: Every marble has a different number, so any single number has a 1 in 8 chance of being picked.

Exam Tip: Since all marbles have different numbers, they are equally likely outcomes with a probability of 1 divided by the total number of items.

Question 7. Two coins are tossed simultaneously 100 times and the outcomes are as given below:

If the same pair of coins is tossed again at random, find the probability of getting :
(i) two heads
(ii) exactly one head
(iii) no head.
Answer:
The total number of coin tosses performed is 100.
(i) The number of times two heads appeared is 21.
\( P(\text{getting two heads}) = \frac{21}{100} \)
(ii) The number of times exactly one head appeared is 55.
\( P(\text{getting exactly one head}) = \frac{55}{100} = \frac{11}{20} \)
(iii) The number of times no head appeared is 24.
\( P(\text{getting no head}) = \frac{24}{100} = \frac{6}{25} \)
In simple words: Look at the table to find the count for each outcome. Divide that count by 100, then simplify.

Exam Tip: Be careful when simplifying your fractions. Always look for common factors like 5 for 55/100 and 4 for 24/100.

Question 8. A bag contains 4 white and 6 black balls,- all of the same shape and same size. A ball is drawn from the bag without looking into the bag. Find the probability that the ball drawn is :
(i) a black ball
(ii) a white ball
(iii) not a black ball
Answer:
The bag holds 4 white balls and 6 black balls.
The sum of all balls inside the bag is \( 4 + 6 = 10 \).
(i) The count of black balls is 6.
\( P(\text{drawing a black ball}) = \frac{6}{10} = \frac{3}{5} \)
(ii) The count of white balls is 4.
\( P(\text{drawing a white ball}) = \frac{4}{10} = \frac{2}{5} \)
(iii) A ball that is not black must be a white ball.
The count of non-black balls is 4.
\( P(\text{drawing a ball that is not black}) = \frac{4}{10} = \frac{2}{5} \)
In simple words: Add the balls to get 10. The probability is the number of favorable balls divided by 10, reduced to the lowest form.

Exam Tip: Notice that "not a black ball" gives the same result as "a white ball" since those are the only two options in the bag.

Question 9. In a single throw of a dice, find the probability of getting a number:
(i) 4
(ii) 6
(iii) greater than 4
Answer:
A standard die has 6 possible outcomes, which are 1, 2, 3, 4, 5, and 6.
(i) The number 4 appears on only 1 face.
\( P(\text{getting a 4}) = \frac{1}{6} \)
(ii) The number 6 appears on only 1 face.
\( P(\text{getting a 6}) = \frac{1}{6} \)
(iii) The numbers on a die that are larger than 4 are 5 and 6, giving 2 favorable options.
\( P(\text{getting a number greater than 4}) = \frac{2}{6} = \frac{1}{3} \)
In simple words: A die has 6 sides. To get one specific number, the chance is 1 in 6. To get 5 or 6, the chance is 2 in 6, which is 1 in 3.

Exam Tip: When dealing with inequalities like "greater than 4", list the valid outcomes first to make sure you do not include 4.

Question 10. Hundred identical cards are numbered from 1 to 100. The cards are well shuffled and then a card is drawn. Find the probability that the number on the card drawn is :
(i) 50
(ii) 80
(iii) 40
Answer:
There is a total of 100 cards.
(i) Only one card contains the number 50.
\( P(\text{drawing card 50}) = \frac{1}{100} \)
(ii) Only one card contains the number 80.
\( P(\text{drawing card 80}) = \frac{1}{100} \)
(iii) Only one card contains the number 40.
\( P(\text{drawing card 40}) = \frac{1}{100} \)
In simple words: Since each card from 1 to 100 is in the deck only once, the chance of picking any specific number is always 1 out of 100.

Exam Tip: For any individual card selection from a set of unique items, the probability of selecting any specific card is always 1 over the total number of cards.

Exercise 22(B)

Question 1. Suppose S is the event that will happen tomorow and P(S) = 0.03.
(i) State in words, the complementary event S’.
(ii) Find P(S’)
Answer:
We are given that \( P(S) = 0.03 \).
(i) The complementary event \( S' \) means that the event will not occur tomorrow.
(ii) We know that the sum of the probabilities of an event and its complement is 1.
\( P(S') = 1 - P(S) \)

\( \implies P(S') = 1 - 0.03 = 0.97 \)
In simple words: The complement of an event is its opposite. Since the total probability is always 1, we subtract 0.03 from 1 to get 0.97.

Exam Tip: Remember the fundamental rule that \( P(S) + P(S') = 1 \). This helps you quickly find complementary probabilities.

Question 2. Five Students A, B, C, D and E are competing in a long distance race. Each student’s probability of winning the race is given below:
A → 20 %, B → 22 %, C → 7 %, D → 15% and E → 36 %
(i) Who is most likely to win the race ?
(ii) Who is least likely to win the race ?
(iii) Find the sum of probabilities given.
(iv) Find the probability that either A or D will win the race.
(v) Let S be the event that B will win the race.
(a) Find P(S)
(b) State, in words, the complementary event S’.
(c) Find P(S’)
Answer:
The probabilities of winning for each student are:
\( P(A) = 20\% \), \( P(B) = 22\% \), \( P(C) = 7\% \), \( P(D) = 15\% \), and \( P(E) = 36\% \).
(i) Student E has the highest probability of winning at 36%, so E is the most likely to win.
(ii) Student C has the lowest probability of winning at 7%, so C is the least likely to win.
(iii) Summing all the given probabilities:
Sum \( = 20\% + 22\% + 7\% + 15\% + 36\% = 100\% \)
(iv) The probability that either A or D wins:
\( P(\text{either A or D wins}) = 20\% + 15\% = 35\% = \frac{35}{100} = \frac{7}{20} \)
(v) (a) The probability that B wins:
\( P(S) = 22\% = \frac{22}{100} = \frac{11}{50} \)
(b) The complementary event \( S' \) means that student B will not win the race.
(c) The probability of the complement is calculated as:
\( P(S') = 1 - P(S) = 1 - \frac{11}{50} = \frac{50 - 11}{50} = \frac{39}{50} \)
In simple words: Compare the percentages to find who has the biggest and smallest chance. For "either A or D", add their percentages. For "not B", subtract B's chance from the total of 1.

Exam Tip: When expressing probabilities, you can use percentages, but it is often expected to simplify them into their fraction forms.

Question 3. A Ticket is randomly selected from a basket containing3 green, 4 yellow and 5 blue tickets. Determine the probability of getting:
(i) a green ticket
(ii) a green or yellow ticket.
(iii) an orange ticket.
Answer:
The basket has 3 green, 4 yellow, and 5 blue tickets.
The total number of tickets is \( 3 + 4 + 5 = 12 \).
(i) Probability of picking a green ticket:
\( P(\text{getting a green ticket}) = \frac{3}{12} = \frac{1}{4} \)
(ii) The total count of green and yellow tickets is \( 3 + 4 = 7 \).
Probability of picking a green or yellow ticket:
\( P(\text{getting a green or yellow ticket}) = \frac{7}{12} \)
(iii) There are no orange tickets in the basket, which means the count is 0.
Probability of picking an orange ticket:
\( P(\text{getting an orange ticket}) = \frac{0}{12} = 0 \)
In simple words: Add all the tickets to get 12. Divide the number of tickets you want by 12, and then simplify the fraction if possible.

Exam Tip: If an outcome is impossible, its probability is always 0. Be sure to show the calculation \( \frac{0}{12} = 0 \) to show your work.

Question 4. Ten cards with numbers 1 to 10 written on them are placed in a bag. A card is chosen from the bag at random. Determine the probability of choosing:
(i) 7
(ii) 9 or 10
(iii) a number greater than 4
(iv) a number less than 6
Answer:
The total number of outcomes is 10, representing the numbers 1, 2, 3, 4, 5, 6, 7, 8, 9, and 10.
(i) Only 1 card is numbered 7.
\( P(\text{getting a number 7}) = \frac{1}{10} \)
(ii) The cards labeled 9 or 10 give 2 favorable options.
\( P(\text{getting 9 or 10}) = \frac{2}{10} = \frac{1}{5} \)
(iii) The numbers larger than 4 are 5, 6, 7, 8, 9, and 10, which are 6 cards.
\( P(\text{getting a number greater than 4}) = \frac{6}{10} = \frac{3}{5} \)
(iv) The numbers smaller than 6 are 1, 2, 3, 4, and 5, which are 5 cards.
\( P(\text{getting a number less than 6}) = \frac{5}{10} = \frac{1}{2} \)
In simple words: Write out the numbers that fit each description. Count how many there are, and divide that by 10.

Exam Tip: Be sure to list the numbers that satisfy the condition (like 5, 6, 7, 8, 9, 10 for "greater than 4") in your steps to avoid counting mistakes.

Question 5. A carton contains eight brown and four white eggs. Find the probability that an egg selected at random is :
(i) brown
(ii) white
Answer:
The carton contains 8 brown eggs and 4 white eggs.
The sum of all eggs is \( 8 + 4 = 12 \).
(i) Probability of picking a brown egg:
\( P(\text{getting a brown egg}) = \frac{8}{12} = \frac{2}{3} \)
(ii) Probability of picking a white egg:
\( P(\text{getting a white egg}) = \frac{4}{12} = \frac{1}{3} \)
In simple words: Add the eggs to get a total of 12. The chance of getting a brown egg is 8 out of 12, and a white egg is 4 out of 12.

Exam Tip: Since all eggs are either brown or white, check that the sum of both probabilities is exactly 1.

Question 6. A box contains 3 yellow, 4 green and 8 blue tickets. A ticket is chosen at random. Find the probability that the ticket is :
(i) yellow
(ii) green
(iii) blue
(iv) red
(v) not yellow
Answer:
The total count of yellow, green, and blue tickets inside the box is:
Total tickets \( = 3 + 4 + 8 = 15 \)
(i) Probability of choosing a yellow ticket:
\( P(\text{getting a yellow ticket}) = \frac{3}{15} = \frac{1}{5} \)
(ii) Probability of choosing a green ticket:
\( P(\text{getting a green ticket}) = \frac{4}{15} \)
(iii) Probability of choosing a blue ticket:
\( P(\text{getting a blue ticket}) = \frac{8}{15} \)
(iv) Since there are no red tickets in the box, the count of favorable tickets is 0.
\( P(\text{getting a red ticket}) = \frac{0}{15} = 0 \)
(v) A ticket that is not yellow must be green or blue.
The sum of green and blue tickets is \( 4 + 8 = 12 \).
\( P(\text{not getting a yellow ticket}) = \frac{12}{15} = \frac{4}{5} \)
Alternatively:
\( P(\text{not getting a yellow ticket}) = 1 - P(\text{getting a yellow ticket}) = 1 - \frac{1}{5} = \frac{4}{5} \)
In simple words: Add all the tickets to get 15. Divide the number of colored tickets by 15. "Not yellow" means green or blue.

Exam Tip: For the "not yellow" part, showing both methods of solving (direct addition and complementary subtraction) demonstrates a thorough understanding of probability concepts.

Question 7. The following table shows number of males and number of females of a small locality in different age groups.

If one of the persons, from this locality, is picked at random, what is the probability that
(a) the person picked is a male ?
(b) the person picked is a female ?
(c) the person picked is a female aged 21-50 ?
(d) the person is a male with age upto 50 years?
Answer:
First, we calculate the total population of the locality:
Total number of males \( = 8 + 12 + 6 = 26 \)
Total number of females \( = 6 + 10 + 4 = 20 \)
Total population \( = 26 + 20 = 46 \)
(a) Probability that the chosen person is male:
\( P(\text{Male}) = \frac{26}{46} = \frac{13}{23} \)
(b) Probability that the chosen person is female:
\( P(\text{Female}) = \frac{20}{46} = \frac{10}{23} \)
(c) The number of females in the 21-50 age group is 10.
Probability that the chosen person is a female aged 21-50:
\( P(\text{Female aged 21-50}) = \frac{10}{46} = \frac{5}{23} \)
(d) The number of males with age up to 50 years includes those in the 10-20 group (8) and the 21-50 group (12):
Total \( = 8 + 12 = 20 \)
Probability that the chosen person is a male aged up to 50 years:
\( P(\text{Male aged up to 50}) = \frac{20}{46} = \frac{10}{23} \)
In simple words: Find the total number of people in the town by adding all males and females together. Divide the number of people matching the category by 46.

Exam Tip: Make sure you carefully identify who falls under "up to 50 years" by adding the 10-20 and 21-50 categories while leaving out the "Above 50" group.