Probability Class 9 questions are provided here as per the latest syllabus of the Class 9 NCERT curriculum. Practicising these Class 9 probability questions will help you score good marks in the Class 9 exams. However, we know that probability plays an important role in our daily existence. Thus, solving different types of questions in probability will help you deal with various real-life situations.

What is the probability?

Probability is the chance of occurrence of an event when we conduct an experiment. The formula for the probability of an event E is given as:

P(E) = Number of outcomes favourable to the event E/Total number of outcomes

Or

P(E) = n(E)/n(S)

Some important points about probability are listed below.

• Probability of an impossible event = 0
• Probability of sure even = 1
• Sum of probabilities of all events associated with a random experiment = 1
• Probability of an event cannot be negative.

Also check: Probability Class 9 Notes

Probability Class 9 Questions and Answers

1. A die is thrown 1000 times with the frequencies for the outcomes 1, 2, 3, 4, 5 and 6 as given in the following table:

Find the probability of getting each outcome.

Solution:

Given,

n(S) = 1000

Let E_i be the event of getting an outcome i.

So, E₁ = Outcome 1

n(E₁) = 179

P(E₁) = n(E₁)/n(S) = 179/1000 = 0.179

Similarly,

P(E₂) = n(E₂)/n(S) = 150/1000 = 0.15

P(E₃) = n(E₃)/n(S) = 157/1000 = 0.157

P(E₄) = n(E₄)/n(S) = 149/1000 = 0.149

P(E₅) = n(E₅)/n(S) = 175/1000 = 0.175

P(E₆) = n(E₆)/n(S) = 190/1000 = 0.19

2. In a survey of 364 children aged 19-36 months, it was found that 91 liked to eat potato chips. If a child is selected at random, find the probability that they do not like to eat potato chips.

Solution:

Given,

Total number of survey children aged from 19-36 months, n(S) = 364

Number of children who liked to eat potato chips = 91

Let E be the event of choosing the number of children who do not like to eat potato chips.

So, n(E) = 364 – 91 = 273

P(E)= n(E)/n(S) = 273/364 = 0.75

Hence, the probability that they do not like to eat potato chips is 0.75.

3. In a cricket match, a batswoman hits a boundary 6 times out of 30 balls she plays. Find the probability that she did not hit a boundary.

Solution:

Given,

Total number of balls played =30

Number of times the batswoman hits a boundary = 6

Thus, the number of times that the batswoman does not hit a boundary = 30 – 6 = 24

P(she did not hit a boundary) = Number of times when she did not hit boundary/Total number of balls played

= 24/30

= 4/5

P(she did not hit a boundary) = 45

4. 80 bulbs are selected at random from a lot, and their lifetime (in hours) is recorded in the form of a frequency table given below.

What is the probability that bulbs selected randomly from the lot have a life of less than 900 h?

Solution:

Given,

Total number of bulbs in a lot = n(S) = 80

Let E be the event of selecting a bulb that has a lifetime of less than 900 hours.

So, n(E) = 10 + 12 + 23 = 45

P(E) = n(E)/n(S) = 45/80 = 9/16

Hence, the probability that the bulb has a lifetime of less than 900 is 9/16.

5. Fifty seeds were selected at random from each of the 5 bags of seeds and were kept under standardised conditions favourable to germination. After 20 days, the number of seeds which had germinated in each collection was counted and recorded as follows:

What is the probability of germination of:

(i) More than 40 seeds in a bag?

(ii) 49 seeds in a bag?

(iii) More than 35 seeds in a bag?

Solution:

From the given,

The total number of bags is 5.

(i) Number of bags in which more than 40 seeds germinated out of 50 seeds = 3

.i.e 41, 42 and 48

P(germination of more than 40 seeds in a bag) = 3/5 = 0.6

(ii) Number of bags in which 49 seeds germinated = 0

P(germination of 49 seeds in a bag) = 0/5 = 0

(iii) Number of bags in which more than 35 seeds germinated = 5

.i.e 40, 48, 42, 39, and 41

So, the required probability = 5/5 = 1

6. Two dice are thrown simultaneously 500 times. Each time the sum of two numbers appearing on their tops is noted and recorded as given in the following table.

If the dice are thrown once more, then what is the probability of getting a sum:

(i) 3?

(ii) More than 10?

(iii) Less than or equal to 5?

(iv) Between 8 and 12?

Solution:

Given that two dice are thrown simultaneously 500 times.

So, n(S) = 500

(i) Getting a sum 3

Let E₁ be the event of getting a sum 3.

That means, n(E₁) = 30

P(E₁) = n(E₁)/n(S) = 30/500 = 0.06

Therefore, the probability of getting a sum equal to 3 = 0.06

(ii) Getting a sum more than 10

Let E₂ be the event of getting a sum more than 10 .i.e sum = 11, 12

So, n(E₂) = 28 + 15 = 43

P(E₂) = n(E₂)/n(S) = 43/500 = 0.086

Therefore, the probability of getting a sum more than 10 = 0.086

(iii) Getting a sum less than or equal to 5

Let E₃ be the event of getting a sum less than or equal to 5 .i.e sum = 2, 3, 4 and 5.

So, n(E₃) = 14 + 30 + 42 + 55 = 141

P(E₃) = n(E₃)/n(S) = 141/500 = 0.282

Therefore, the probability of getting a sum less than or equal to 5 = 0.282

(iv) Getting a sum between 8 and 12

Let E₄ be the event of getting a sum between 8 and 12 .i.e sum = 9, 10, 11.

n(E₄) = 53 + 46 + 28 = 127

P(E₄) = n(E₄)/n(S) = 127/500 = 0.254

Therefore, the probability of getting a sum between 8 and 12 = 0.254

7. Three coins are tossed simultaneously 200 times with the following frequencies of different outcomes:

If the three coins are simultaneously tossed again, compute the probability of 2 heads coming up.

Solution:

Given that three coins are tossed simultaneously 200 times.

So, n(S) = 200

Let A be the event of getting 2 heads.

n(E) = 72

P(E) = n(E)/n(S) = 72/200 = 0.36

Therefore, the probability of 2 heads coming up is 0.36.

8. Can the experimental probability of an event be a negative number? If not, why?

Solution:

The experimental probability of an event cannot be a negative number, and it is always a positive integer since the number of trials in which the event can happen is a positive number. Thus, the experimental probability is always positive.

9. Over the past 200 working days, the number of defective parts produced by a machine is given in the following table.

Determine the probability that tomorrow’s output will have:

(i) No defective part.

(ii) At least one defective part.

(iii) Not more than 5 defective parts.

(iv) More than 13 defective parts.

Solution:

Given,

Total number of working days = 200

n(S) = 200

(i) No defective part

Let A be the event of getting output with no defective part.

n(A) = 50

P(A) = n(A)/n(S) = 50/200 = 0.25

Therefore, the probability that tomorrow’s output will have no defective part is 0.25.

(ii) At least one defective part

Let B be the event getting output with at least one defective part.

Thus, n(B) = Total number of days – Number of days with zero defective parts

= 200 – 50

= 150

P(B) = n(B)/n(S) = 150/200 = 0.75

Therefore, the probability that tomorrow’s output will have at least one defective part is 0.75.

(iii) Not more than 5 defective parts

Let C be the event of getting output with not more than 5 defective parts.

So, n(C) = 50 + 32 + 22 + 18 + 12 + 12 = 146

P(C) = n(C)/n(S) = 146/200 = 0.73

Therefore, the probability that tomorrow’s output with not more than 5 defective parts is 0.73.

(iv) More than 13 defective parts

Let D be the event of getting output with more than 13 defective parts.

However, there is no entry with more than 13 defective parts.

So, n(D) = 0

P(D) = n(D)/n(E) = 0/200 = 0

Hence, the probability that tomorrow’s output with more than 13 defective parts is 0.

10. Eleven bags of wheat flour, each marked 5 kg, actually contained the following weights of flour (in kg):

4.97 5.05 5.08 5.03 5.00 5.06 5.08 4.98 5.04 5.07 5.00

Find the probability that any of these bags chosen at random contains more than 5 kg of flour.

Solution:

Given,

Total number of bags = 11

.i.e n(S) = 11

Let E be the event of choosing a bag that contains more than 5 kg of flour.

E = {5.05, 5.08, 5.03, 5.06, 5.08, 5.04, 5.07}

n(E) = 7

P(E) = n(E)/n(S) = 7/11

Therefore, the probability that the chosen bag contains more than 5 kg of flour is 7/11.

Practice Questions for Probability Class 9

1. In a sample study of 642 people, it was found that 514 people have a high school certificate. If a person is selected randomly, find the probability that the person has a high school certificate.
2. Bulbs are packed in cartons, each containing 40 bulbs. Seven hundred cartons were examined for defective bulbs, and the results are given in the following table:

Number of defective bulbs	0	1	2	3	4	5	6	More than 6
Frequency	400	180	48	41	18	8	3	2

One carton was selected at random. What is the probability that it has:

(i) No defective bulb?

(ii) Defective bulbs from 2-6?

(iii) Defective bulbs less than 4?

3. To know the opinion of the students about the subject statistics, a survey of 200 students was conducted. The data is recorded in the following table.

Opinion	Like	Dislike
Number of students	135	65

Find the probability that a student is chosen at random (i) Likes statistics, (ii) Does not like it.

4. The percentage of marks obtained by a student in the monthly unit tests is given below.

Unit test	I	II	III	IV	V
Percentage of marks obtained	69	71	73	68	74

What is the probability that the student gets more than 70% marks in a unit test?

5. Two coins are tossed simultaneously 500 times, and we get two heads:

Find the probability of occurrence of each of these events.