CBSE Class 9 Mathematics Probability Set 03

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

Section - A

Question 1. Define an event.
Answer: An event is a collection or group of outcomes of an experiment, denoted by capital letters of the English alphabet such as \(A\), \(B\), \(E\), etc.
In simple words: An event is a specific set of results or outcomes from an experiment.

Exam Tip: Remember that events are typically named with uppercase letters.

Question 2. Give definition of probability.
Answer: The empirical (or experimental) probability \(P(E)\) of an event \(E\) is given by the ratio of the number of trials in which the event has happened to the total number of trials:
\(P(E) = \frac{\text{Number of trials in which } E \text{ has happened}}{\text{Total number of trials}}\)
In simple words: Probability measures how likely an event is to happen based on observations or experiments.

Exam Tip: Always write the formula clearly with numerator and denominator before substituting values.

Question 3. Probability of certain event is …………
Answer: 1
In simple words: An event that is certain to happen always has a probability of 1.

Exam Tip: A probability of 1 means 100 percent certainty.

Question 4. Probability of impossible event is ………..
Answer: 0
In simple words: An event that can never happen has a probability of 0.

Exam Tip: Probabilities can never be negative or greater than 1.

Question 5. Which is not a probability of an event?
(a) 2
(b) \(\frac{2}{3}\)
(c) .001
(d) .25
Answer: (a) 2
The probability of any event always lies between 0 and 1 inclusive. Since 2 is greater than 1, it cannot be a probability.
In simple words: Probability values must always be between 0 and 1.

Exam Tip: Always check if a given value is between 0 and 1 (inclusive) when identifying valid probabilities.

Question 6. A bag contains 50 coins and each coin marked from 51 to 100. One coin is picked up at random. The probability that the number on the coin is not a prime number is……………………
Answer: \(\frac{4}{5}\)
Total number of coins = 50 (numbered from 51 to 100).
Prime numbers between 51 and 100 are: 53, 59, 61, 67, 71, 73, 79, 83, 89, 97 (Total 10 prime numbers).
Number of coins with non-prime numbers = \(50 - 10 = 40\).
Probability of getting a non-prime number = \(\frac{40}{50} = \frac{4}{5}\).
In simple words: Subtract the count of prime numbers from the total coins to find non-prime coins, then divide by the total number of coins.

Exam Tip: Be careful when counting prime numbers in a given range; list them out to avoid missing any.

Section - B

Question 7. A coin is tossed 1000 times with the following frequencies. Head: 455, Tail: 545, compute the probability for each event.
Answer:
Total number of trials = 1000
Number of times Head appears = 455
Number of times Tail appears = 545
Probability of getting a Head \(P(\text{Head}) = \frac{455}{1000} = 0.455\)
Probability of getting a Tail \(P(\text{Tail}) = \frac{545}{1000} = 0.545\)
In simple words: Divide the frequency of each outcome by the total number of coin tosses.

Exam Tip: The sum of probabilities of all possible outcomes of an event is always 1 (\(0.455 + 0.545 = 1\)).

Question 8. In a cricket match, a batsman hits a boundary 6 times out of 30 balls plays. Find the probability that on a ball played. (i) He hits boundary (ii) He does not ht a boundary.
Answer:
Total number of balls played = 30
Number of times he hits a boundary = 6
Number of times he does not hit a boundary = \(30 - 6 = 24\)
(i) Probability that he hits a boundary = \(\frac{6}{30} = \frac{1}{5} = 0.2\)
(ii) Probability that he does not hit a boundary = \(\frac{24}{30} = \frac{4}{5} = 0.8\)
In simple words: Find successful outcomes and unsuccessful outcomes, then divide each by the total number of balls.

Exam Tip: You can also find the second probability by subtracting the first probability from 1: \(1 - 0.2 = 0.8\).

Question 9. Three coins tossed simultaneously 100 times with the following frequencies of different outcomes.
Outcome: No head, One head, Two head, Three head
Frequency: 14, 38, 36, 12
If the coin tossed again then find the probability.
(i) two heads coming up
(ii) 3 heads coming up
(iii) getting more tails than heads
(iv) at least one head coming up
Answer:
Total trials = 100
(i) Frequency of two heads = 36
\(P(\text{two heads}) = \frac{36}{100} = 0.36\)
(ii) Frequency of three heads = 12
\(P(\text{three heads}) = \frac{12}{100} = 0.12\)
(iii) Getting more tails than heads means getting either 0 heads (3 tails) or 1 head (2 tails).
Frequency (0 heads or 1 head) = \(14 + 38 = 52\)
\(P(\text{more tails than heads}) = \frac{52}{100} = 0.52\)
(iv) At least one head coming up means 1 head, 2 heads, or 3 heads.
Frequency = \(38 + 36 + 12 = 86\)
\(P(\text{at least one head}) = \frac{86}{100} = 0.86\)
In simple words: Sum the frequencies of the required outcomes and divide by the total frequency of 100.

Exam Tip: "At least one" means total outcomes minus the outcome with zero occurrences (\(100 - 14 = 86\)).

Question 10. In a football match, a player makes 4 goals from 10 kicks. The probability of a goal is from 10 kicks is.
Answer:
Total number of kicks = 10
Number of goals scored = 4
Probability of scoring a goal = \(\frac{4}{10}\)
In simple words: Divide the number of successful goals by the total number of kicks.

Question 11. The percentage of marks obtained by a student in the monthly unit tests are given as :
Unit Test: I, II, III, IV, V
% marks obtained: 58, 64, 76, 62, 85
Find the probability that the student get a distinction (marks more than 75%)
Answer:
Total number of unit tests = 5
Tests with marks more than 75% (distinction): III (76%) and V (85%) = 2 tests.
Probability of getting a distinction = \(\frac{2}{5} = 0.4\)
In simple words: Count how many tests had marks above 75 and divide by the total number of tests.

Exam Tip: Carefully check the condition "more than 75%" (75% itself is not included, though none scored exactly 75 here).

Question 12. 1000 families with 2 children were selected randomly, and the following data were recorded.
No. of boys in a family: 0, 1, 2
No. of families: 140, 560, 300
If a family chosen at random, find the prob. that it has
(a) No boys
(b) One boy
(c) Two boys
(d) at least one boy
(e) at most two boy.
Answer:
Total number of families = 1000
(a) Families with no boys = 140
\(P(\text{no boys}) = \frac{140}{1000} = 0.14\)
(b) Families with one boy = 560
\(P(\text{one boy}) = \frac{560}{1000} = 0.56\)
(c) Families with two boys = 300
\(P(\text{two boys}) = \frac{300}{1000} = 0.30\)
(d) Families with at least one boy (1 or 2 boys) = \(560 + 300 = 860\)
\(P(\text{at least one boy}) = \frac{860}{1000} = 0.86\)
(e) Families with at most two boys (0, 1, or 2 boys) = \(140 + 560 + 300 = 1000\)
\(P(\text{at most two boys}) = \frac{1000}{1000} = 1\)
In simple words: Match the required condition to the frequency table and divide by 1000.

Exam Tip: "At most two" means 2 or fewer, which covers all families in this dataset, making it a certain event with probability 1.

Question 13. The record of a weather station shows that out of the past 250 consecutive days, its weather forecast correct 175 times. What is the probability that on a given day. (i) it was correct. (ii) it was not correct.
Answer:
Total number of days = 250
Number of correct forecasts = 175
Number of incorrect forecasts = \(250 - 175 = 75\)
(i) Probability that it was correct = \(\frac{175}{250} = \frac{7}{10} = 0.7\)
(ii) Probability that it was not correct = \(\frac{75}{250} = \frac{3}{10} = 0.3\)
In simple words: Divide the number of correct and incorrect days separately by the total number of recorded days.

Exam Tip: Ensure your decimal conversions match standard fractional simplifications.

Section - D

Question 14. A die is thrown 1000 times with following frequency of outcomes 1, 2, 3, 4, 5 and 6 as given below
No. on die: 1, 2, 3, 4, 5, 6
Frequency: 179, 150, 157, 149, 175, 190
Find the probability of each outcome.
Answer:
Total number of trials = 1000
(i) \(P(\text{outcome 1}) = \frac{179}{1000} = 0.179\)
(ii) \(P(\text{outcome 2}) = \frac{150}{1000} = 0.150 = 0.15\)
(iii) \(P(\text{outcome 3}) = \frac{157}{1000} = 0.157\)
(iv) \(P(\text{outcome 4}) = \frac{149}{1000} = 0.149\)
(v) \(P(\text{outcome 5}) = \frac{175}{1000} = 0.175\)
(vi) \(P(\text{outcome 6}) = \frac{190}{1000} = 0.190 = 0.19\)
In simple words: Divide each frequency by the total number of throws (1000).

Exam Tip: Trailing zeros in decimals can be dropped (e.g., 0.150 becomes 0.15).

Question 15. Following table shows the marks scored by a group of 90 students in a mathematics test of 100 marks.
Marks: 0-20, 20-30, 30-40, 40-50, 50-60, 60-70, 70-80
No. of student: 7, 10, 10, 20, 20, 15, 8
Find the probability that a student obtained
(i) less than 20% marks (ii) 60 or more marks
Answer:
Total number of students = 90
(i) Students obtaining less than 20% marks (marks in interval 0-20) = 7
\(P(\text{less than 20\% marks}) = \frac{7}{90}\)
(ii) Students obtaining 60 or more marks (marks in intervals 60-70 and 70-80) = \(15 + 8 = 23\)
\(P(\text{60 or more marks}) = \frac{23}{90}\)
In simple words: Add the student counts for the required mark intervals and divide by 90.

Exam Tip: Be careful to include all relevant class intervals when grouping data.

Question 16. The following table gives the life of 400 lamps.
Life time in Hours: 300-400, 400-500, 500-600, 600-700, 700-800, 800-900, 900-1000
No. of Lamp: 14, 56, 60, 86, 74, 62, 48
A bulb is selected at random find the probability that the life time of the selected bulb is:
(i) less than 400
(ii) between 300 to 800 hours.
(iii) at least 700 hours.
Answer:
Total number of lamps = 400
(i) Lamps with lifetime less than 400 hours (interval 300-400) = 14
\(P(\text{less than 400}) = \frac{14}{400} = \frac{7}{200}\)
(ii) Lamps with lifetime between 300 to 800 hours (intervals 300-400, 400-500, 500-600, 600-700, 700-800) = \(14 + 56 + 60 + 86 + 74 = 290\)
\(P(\text{between 300 to 800}) = \frac{290}{400} = \frac{29}{40}\)
(iii) Lamps with lifetime at least 700 hours (intervals 700-800, 800-900, 900-1000) = \(74 + 62 + 48 = 184\)
\(P(\text{at least 700}) = \frac{184}{400} = \frac{23}{50}\)
In simple words: Sum the frequencies for the specified lamp life ranges and simplify the resulting fraction.

Exam Tip: Always reduce fractions to their simplest form in final answers.

Question 17. The percentage of attendance of different classes in a year in a school is given below:
Class: X, IX, VIII, VII, VI, V
Attendance: 30, 62, 85, 92, 76, 55
What is the probability that the class attendance is more than 75%
Answer:
Total number of classes = 6
Classes with attendance more than 75% are: VIII (85%), VII (92%), and VI (76%) = 3 classes.
Probability = \(\frac{3}{6} = \frac{1}{2}\)
In simple words: Count how many classes have attendance strictly greater than 75 and divide by the total number of classes (6).

Exam Tip: Count the total number of categories in the denominator, not the sum of percentages.