Class 9 - Maths - Probability

**Exercise 15.1**

**Question1:**

In an ODI match, a wicket keeper drops a catch 6 times out of 30 catches he gets. Find the probability of the wicketkeeper not dropping a catch.

Answer:

Let E be the event of dropping a catch

P(E) = Probability of the wicketkeeper dropping a catch = 6/30 = 1/5

Thus, Probability of not dropping a catch is 1 - P(E) = 1 – 1/5 = 4/5

**Question 2:**

1500 families with 2 children were selected randomly, and the following data were recorded:

What is the probability that a family, chosen at random, has

- 2 girls ii. 1 girl iii. No girl

Also check whether the sum of these probabilities is 1.

Answer:

Total number of families = 475 + 814 + 211 = 1500

- P(Probability of 2 girls) = No. of family having 2 girls/total no. of girls

= 475/1500

- P(Probability of 1 girl) = No. of family having 1 girl/total no. of girls

= 814/1500

iii. P(Probability of No girl) = No. of family having no girl/total no. of girls

= 211/1500

Sum of all Probabilities = P(Probability of 2 girls) + P(Probability of 1 girl) + P(Probability of No

girl)

= 475/1500 + 814/1500 + 211/1500

= (475 + 814 + 211)/1500

= 1500/1500

= 1

Hence, the sum of these probabilities is 1.

**Question 3:**

Refer to Example 5, Section 14.4, Chapter 14. Find the probability that a student of the class was born in August.

Answer:

Total number of students = 40

Number of students born in August = 6

P( Student born in August) = Number of students born in August/Total number of students

= 6/40

= 3/20

**Question4:**

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.

Answer:

Total number of times the coins were tossed = 200

Number of times 2 heads occur = 72

So, P(2 heads coming up) = 72/200 = 9/25

**Question 5:**

An organization selected 2400 families at random and surveyed them to determine a relationship between income level and the number of vehicles in a family.

The information gathered is listed in the table below:

Suppose a family is chosen. Find the probability that the family chosen is

(i) earning Rs 10000 – 13000 per month and owning exactly 2 vehicles.

(ii) earning Rs 16000 or more per month and owning exactly 1 vehicle.

(iii) earning less than Rs 7000 per month and does not own any vehicle.

(iv) earning Rs 13000 – 16000 per month and owning more than 2 vehicles.

(v) owning not more than 1 vehicle.

Answer:

Total number of families: 2400

(i) P( Earning Rs 10000-13000 per month and owning exactly 2 vehicle) = 29/2400

(ii) P( Earning Rs 16000 or more per month and owning exactly 1 vehicle) = 579/2400

= 193/800

(iii) P( Earning less than Rs 7000 per month and does not own any vehicle) = 10/2400

= 1/240

(iv) P(Earning Rs 13000 – 16000 per month and owning more than 2 vehicles) = 25/2400

= 1/96

(v) P(owning not more than 1 vehicle) = P(a family owning 0 vehicle or 1 vehicle)

= (10 + 0 + 1 + 2 + 1 + 160 + 305 + 535 + 469 + 579)/2400

= 2062/2400 = 1031/1200

**Question 6:**

Refer to Table 14.7, Chapter 14.

(i) Find the probability that a student obtained less than 20% in the mathematics test.

(ii) Find the probability that a student obtained marks 60 or above

Answer:

Total number of students = 90

(i) P(a student obtained less than 20%)

= No. of student obtained less than 20%/Total no of students

= 7/90

(ii) P(a student obtained 60 marks or above)

= No. of students who obtained 60 marks or more/Total no. of students

= (15 + 8)/90

= 23/90

**Question 7:**

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.

Find the probability that a student chosen at random

(i) likes statistics (ii) does not like it

Answer:

Total number of students = 200

Number of students liking the subject = 135

Number of students disliking the subject = 65

- P(liking the subject) = 135/200 = 27/40
- P(Disliking the subject) = 65/200 = 13/40

**Question 8:**

Refer to Q.2, Exercise 14.2. What is the empirical probability that an engineer lives:

(i) less than 7 km from her place of work?

(ii) more than or equal to 7 km from her place of work?

(iii) within 1/2 km from her place of work?

Answer:

Total number of engineers = 40

Let us arrange the data in ascending order as follows:

2, 2, 3, 3, 3, 5, 5, 6, 6, 7, 7, 7, 7, 8, 9,9, 10, 10, 11, 11, 12, 12, 12, 12, 12, 13, 14, 15, 15, 16, 17,

17, 18, 18, 19, 20, 25, 31, 32

(i) P(an engineer lives less than 7 km from her place of work)

= Number of engineer lives less than 7 km from her place of work/Total no. of engineer

= 9/40

(ii) P(an engineer lives more than or equal to 7 km from their place of work)

= engineer lives more than or equal to 7 km from their place of work /Total no. of engineer

= 31/40

(iii) P(an engineer lives within 1/2 km from her place of work)

= Number of engineer lives within 1/2 km from her place of work/Total no. of engineer

= 0/40 = 0

Note: Questions 9 and 10 are activities, so student should perform it on their own.

**Question 11:**

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.

Answer:

Total number of bags examined = 11

P(a bag weighing more than 5 kg) = No. of bags which weigh more than 5 kg/Total no. of bags

= 7/11

**Question 12:**

In Q.5, Exercise 14.2, you were asked to prepare a frequency distribution table, regarding the concentration of sulphur dioxide in the air in parts per million of a certain city for 30 days.

Using this table, find the probability of the concentration of sulphur dioxide in the interval 0.12 - 0.16 on any of these days.

Answer:

Total number of students = 30

P(concentration of sulphur dioxide in the interval 0.12 - 0.16 in a day)

= Number of days on which the concentration was in the interval 0.12 - 0.16/Total no. of days

= 2/30

= 1/15

**Question 13:**

In Q.1, Exercise 14.2, you were asked to prepare a frequency distribution table regarding the blood groups of 30 students of a class.

Use this table to determine the probability that a student of this class, selected at random, has blood group AB.

A, B, O, O, AB, O, A, O, B, A, O, B, A, O, O,

A, AB, O, A, A, O, O, AB, B, A, O, B, A, B, O.

Answer:

Total number of students = 30

P(a student has blood group AB)

= No. of students which have the blood group AB/Total no. of students

= 3/30

= 1/10

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