Probability – Exercise 15.1 – Class IX

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

Solution:

Total number of blls played by the batswoman = 30

Boundaries hit = 6

No. of balls in which she did not hit any numbers = 30 – 6 = 24

⸫P(she did not hit a boundary) =  no. of balls in which she did not hit any  boundary/total number of balls played = 24/30 = 4/5

1. 1500 families with 2 children were selected randomly, and the following data were recorded:
 Number of girls in a family 2 1 0 Number  of  families 475 814 211

Compute the probability of a family, chosen at random, having

(i) 2 girls (ii) 1 girl (iii) No girl

Also check whether the sum of these probabilities is 1.

Solution:

(i) P(a family having 2 girls) = no. of families having 2 girls/total no. of families  = 475/1500 = 19/60­

(ii) P ( a family having 1 girl) = no. of families having 2 girls/total no. of families  = 814/1500 = 407/750

(ii) P ( a family having no girl) = no. of families having 2 girls/total no. of families  = 211/1500

Sum of  the probabilities in all three cases  = 19/60 + 407/750 +211/1500   = 475+814+211/1500 = 1500/1500 = 1

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

Total number of students considered = 40

No. of students born in August = 6

⸫ P ( a student born in August) = no. of students born in August/total no. of students considered  = 6/40 = 3/20

1. 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:

Total number of tosses = 200

No. of times 2 heads occur = 72

⸫ P ( two heads coming up) = no. of  times two heads occur/total no. of tosses  = 72/200 = 9/25

1. An organisation 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:
 Monthly income (in Rs.) Vehicles per family 0 1 2 Above 2 Less than 7000 10 160 5 0 7000 – 10000 0 305 27 2 10000 – 13000 1 535 29 1 13000 – 16000 2 469 59 25 16000 or more 1 579 82 88

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.

Solution:

Total no of families considered = 2400

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

= no. of families earning Rs. 10000 – 13000 per month and owning exactly 2 vehicles/total no. of families

= 29/2400

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

= no. of families earning Rs. 16000 or more per month and owning exactly 1 vehicle/total no. of families

= 579/2400 = 193/800

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

= no. of families earning Rs. 7000 per month and does not own any vehicle/total no. of families

= 10/2400 = 1/240

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

= no. of families earning Rs. 13000 – 16000 per month and owning more than 2 vehicles/total no. of families  = 25/2400 = 1/96

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

= 10+0+1+2+1+160+305+535+469+579/2400 = 2062/2400 = 1031/1200

1. 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.

 Marks Number of students 0 – 20 7 20 -30 10 30 – 40 10 40 – 50 20 50 – 60 20 60 – 70 15 70 – above 8 Total 90

Solution:

(i) Total number of students = 90

P(a student obtained less than 20%) = no. of students who obtained less than 20%/total number of students = 7/90

(ii) P(a student obtained 60 marks or above) = no. of students who obtained 60 makrs or more /total numner of students  = 15+8/90 = 23/90

1. 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 Number of students like 135 dislike 65

Find the probability that a student chosen at random

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

Solution:

(i) P(a student like statistics) = no. of students who like statistics/total number of students = 135/200 = 27/40

(i) P(a student does not like statistics) = no. of students who does not like statistics/total number of students = 65/200 = 13/40

1. Refer to Q.2, Exercise 14.2.

5      3     10   2     25   11   13   7     12   31

19    10   12   17   18   11   32   17   16   2

7      9     7     8     3     5     12   15   18   3

12    14   2     9     6     15   15   7     6     12

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?

Solution:

Total no. 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, 15, 16, 17, 17, 18, 18, 19, 20, 25, 31, 32

(i)P(an engineer lives less than 7km from her place to work) = no. of engineers live less than 7km from her place to work/total no. of engineers = 9/40

(ii)P(an engineer lives more than or 7km from her place to work) = no. of engineers live more than or  7km from her place to work/total no. of engineers = 31/40

(iii)P(an engineer lives within 1/2 km from her place to work) = no. of engineers live within 1/2 km from her place to work/total no. of engineers = 0/40 = 0