NEXT

## Chapter 15: Probability

TEXTBOOK’S EXERCISE – 15.1

1. 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.

Sol. Total number of balls played by the batswoman = 30, Boundaries hit = 6

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

∴ P (she did not hit a boundary)

= 24/30 = 4/5

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

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.

Sol. (i) P (a family having 2 girls) =

= 475/ 1500 = 19/60

(ii) P (a family having 1 girl

No. of families having 1 girl/Total no. of families

= 814/ 1500 = 407/750

(iii) P (a family having no girl) = No. of families having no girl/Total no. of families = 211/1500

Sum of the probabilities in all three cases

3. In a particular section of Class IX, 40 students were asked about the months of their birth and the fol­lowing graph was prepared for the data so obtained. Find the probability that a student of the class was born in August.

Sol. Total number of students considered = 40

No. of students born in August = 6

∴ P (a student was born in August) = No. of students born in August Total no. of students considered = 6/40 = 3/20

4. 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.

Sol. Total number of tosses = 200

No. of times 2 heads occur = 72

∴ P (2 heads coming up) = No. of times 2 heads occur/Total no. of tosses = 72/200 = 9/25

5. 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:

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

Sol. 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 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 1 vehicle/Total no. of families = 579/2400

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

= No. of families earning less than 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 owing 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 owning 0 vehicles or 1 vehicle)

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