CBSE Class 9 Mathematics Probability Set C

Very Short Answer Type Questions

Question. The following data represents the number of girls in a family.

A family is chosen at random. Find the probability of having exactly 2 girls in the chosen family.
Answer : Total number of families = 1000 = n(S)
Number of families having exactly 2 girls = 475 = n(E)
∴ P (E) = n(E)/n(S) = 475/1000 = 19/40

Question. The following table shows the birth months of 48 babies in a hospital:

Find the probability of months in which 6 babies were born.
Answer : Number of months = 12 ⇒ n(S) = 12
Let E be the event having months in which 6 babies were born i.e., July, Aug and Dec
⇒ n(E) = 3
∴ P (E) = n(E)/n(S) = 3/12 = 14/5.

Question. A coin is tossed 500 times with the following frequencies : Head : 255, Tail : 245.
Then find the sum of the probabilities of each event.
Answer : P(Head appears) = 255/500 = 51/100
P(Tail appears) = 245/500 = 49/100
Sum of the probabilities = 51/100 + 49/100 = 100/100 = 1

Question. A bag contains 6 green and 5 blue balls. If probability of choosing a green ball randomly is n/11, then find the value of n.
Answer : Number of balls = 6 + 5 = 11 = n(S)
Let E be the event of choosing a green ball randomly.
∴ P (E) = n(E)/n(S) = 6/11 = n/11 ⇒ n = 6

Question. In a single throw of two dice, find the probability that there will be a doublet.
Answer : Number of elements in sample space when two dice are thrown = 6 × 6 = 36
Doublets are {(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)}
i.e., 6 in number
∴ P(getting doublet)
= Number of doublets/Number of elements in sample space = 6/36 = 1/6

Question. A survey of 100 children of a locality shows their favourite sport

Out of these children, one is chosen at random.
What is the probability that the chosen child likes football?
Answer : Total number of students, n(S) = 100
Let E be the event that child likes football.
i.e., n(E) = 48
∴ P (E) = 48/100 = 12/25

Question. In a game, a woman wins 16 times out of 20 balls she plays. Find the probability that she does not win the game.
Answer : P(woman wins the game) = 16/20 = 4/5
P(woman does not win the game) = 1 − 4/5 = 1/5

Question. Ram and Priya are playing a game. Ram’s winning probability is 1/3 . Find the Priya’s winning probability.
Answer : Ram’s winning probability = 1/3
⇒ Priya’s winning probability = 1 − 1/3 = 2/3

Question. Find the probability of choosing a vowel randomly from the letters of the word ‘EXAMINATION’.
Answer : Number of letters in the word ′EXAMINATION′ = 11
Number of vowels in the word = 6
Required probability = 6/11

Question. Based on the given information, find the probability of people with age 60, 61 & 64 who can drive.

Answer : Number of people with age 60, 61 and 64 who can drive = 16090 + 11490 + 3607 = 31187
Total number of people who can drive
= 16090 + 11490 + 8012 + 5448 + 3607 + 2320 = 46967
∴ Required probability = 31187/46967

Short Answer Type Questions

Question. In a locality of 5000 families a survey was conducted and the following data was collected.

Out of these families, a family is chosen at random. What is the probability that the chosen family has less than 5 members?
Answer : Total number of families, n(S) = 5000
Let E be the event that the chosen family has less than 5 members.
i.e., n(E) = 1060 + 1000 + 1020 = 3080
∴ P(E) = n(E)/n(S) = 3080/5000 = 77/125

Question. In 60 throws of a die, the outcomes were noted as below:

If die is thrown at random, then what is the probability that upper face of a die shows an even prime number? Also find the probability that upper face shows an odd number.
Answer : Total number of throws, n(S) = 60
Let E be the event that upper face shows an even prime number, i.e.,
n(E) = 10 [∵ 2 is the only even prime number]
∴ P(E) = n(E)/n(S) = 10/60 = 1/6
Let F be the event that upper face shows an odd number
= 8 + 15 + 7 = 30
∴ P(F) = n(F)/n(s) = 30/60 = 1/2

Question. An integer is chosen at random from the first 200 positive integers. Find the probability that the integer is divisible by 11.
Answer : Total number of integers in the sample space
= 200 = n(S).
Among first 200 positive integers, we have 11, 22, 33, 44, 55, 66, 77, 88, 99, 110, 121, 132, 143, 154, 165, 176, 187, 198 are divisible by 11.
∴ Number of integers which are divisible by 11 = 18 P(integer is divisible by 11)
= Number of integers divisible by 11/Total number of integers in sample space
= 18/200 = 9/100

Question. A die is thrown 100 times and following observations were recorded:

Find the probability that the die shows
(i) a number less than 3.
(ii) a number greater than 4.
Answer : n(S) = 100
(i) Let E be the event that the die shows a number less than 3, i.e., n(E) = 12 + 18 = 30
∴ P(E) = n(E)/n(S) = 30/100 = 3/10
(ii) Let F be the event that the die shows a number greater than 4, i.e., n(F) = 14 + 16 = 30
∴ P (F) = n(F)/n(S) = 30/100 =  3/10

Question. A die was rolled 100 times and the number of times, 6 came up was noted. If the experimental probability calculated from this information is 2/5, then how many times 6 came up?
Answer : Here, total number of trials = 100
Let x be the number of times 6 came up.
We know, probability of an event
= Frequency of the event occuring/Total number of trials
⇒ x/100 = 2/5
⇒ x = 40

Question. The number of hours spent by Ashu, a school student, on various activities on a working day are given below:

His friend Sonu came to his house to meet Ashu.
What is the probability that
(i) Ashu is available at home.
(ii) Ashu will play outdoor games
Answer : Total number of hours = 24
(i) Number of hours during which Ashu is at home
= 7 + 2 + 2 = 11
∴ Probability that Ashu is available at home = 11/24
(ii) Number of hours during which Ashu plays outdoor games = 3
∴ Probability that Ashu will play outdoor games
= 3/24 = 1/8

Question. The probability of guessing the correct answer to a certain question is x/5. If the probability of not guessing the correct answer is 2x/3. Then, find the value of 26x.
Answer : We have
P(guessing correct answer) = x/5 and P(not guessing correct answer) = 2x/3
Clearly, P(guessing correct answer) + P(not guessing
correct answer) = 1             [ ∵ P(E) + P (not E) = 1]
x/5 + 2x/3 = 1
⇒ 3x + 10x = 15 ⇒ x = 15/13
⇒ 26x = 26 x 15/13 = 30

Question. There are 35 students in class IX–A, 34 in IX-B and 33 in IX–C. Some of them are allotted project on Chapter 2 (Polynomials) and some on Chapter-1 (Number system) as shown in the table.

Find the probability that the student chosen at random,
(i) prepares project on chapter 1
(ii) prepares project on chapter 2
Answer : Total number of students = 35 + 34 + 33 = 102
(i) Number of students prepare project on chapter-1 = 74
∴ Probability that the student prepares project on chapter-1 = 74/102 = 37/51
(ii) Number of students prepare project on chapter-2 = 28
∴  Probability that the student prepares project on chapter-2 = 28/102 = 14/51

Question. The percentage of attendance of different classes in a year in a school is given below:

(i) What is the probability that the class attendance is more than 75%?
(ii) Find the probability that the class attendance is less than 50%.
Answer : Total number of classes = 6
(i) Number of classes in which attendance percentage is more than 75% = 3
∴  Required probability = 3/6 = 1/2
(ii) Number of classes in which attendance is less than 50% = 1
∴  Required probability = 1/6

Question. If the difference between the probability of success and failure (i.e., not success) of an event is 5/19 (assuming probability of failure is greater than that of success). Find the probability of success and failure of the event respectively.
Answer : Let the probability of success be x
Then, probability of failure = 1 – probability of success
⇒ Probability of failure = 1 – x
According to question, we have,
Probability of failure – Probability of success = 5/19
⇒ 1 - x - x = 5/19 ⇒ 1 - 2x = 5/19
⇒ 2x = 1 - 5/19 = 14/19 ⇒ x = 7/19
∴ Probability of success = 7/19 and probability of failure = 1 − 7/19 = 12/19

Click on link below to download CBSE Class 9 Mathematics Probability Set C