9th Maths

Probability

Exercise 15.1

Question 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.

Answer: Number of boundaries = 6
Number of balls = 30
P(E) = (Number of favourable outcomes)/(Total number of event)
`= 6/30 = 1/5`

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

Number of girls in a family210
Number of families475814211



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

(i) 2 girls

Answer: Number of families with 2 girls = 475
Total number of families = 1500
`P(E) = (Num\be\r\  of  \fa\vo\ur\ab\l\e\  \ou\tc\om\es)/(To\ta\l\  \n\u\mb\er\  of\  ev\en\t)`
`= 475/1500 = 19/60`

(ii) 1 girl

Answer: Number of families with 1 girl = 814
P(E) = (Number of favourable outcomes)/(Total number of event)
`= 814/1500 = 407/750`

(iii) No girl

Answer: Number of families with no girl = 211
P(E) = (Number of favourable outcomes)/(Total number of event)
`= 211/1500`


Question 3: Refer to Example 5, Section 14.4, Chapter 14(NCERT Book). 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(E) = (Number of favourable outcomes)/(Total number of event)
`= 6/40 = 3/20`

Question 4: Three coins are tossed simultaneously 200 times with the following frequencies of different outcomes:

Outcome3 heads2 heads1 headNo head
Frequency23727728

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

Answer: Total number of toss = 200
Number of times 2 heads come = 72
P(E) = (Number of favourable outcomes)/(Total number of event)
`= 72/200 = 9/25`


Question 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:

Monthly Income (in Rs. Vehicles per family012Above 2
Less than 700010160250
7000 – 100000305272
10000 – 130001535291
13000 – 1600024695925
16000 or more15798288

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.

Answer: Total number of families = 2400
Number of families with income Rs. 10000 – 13000 per month and owning exactly 2 vehicles = 29
P(E) = (Number of favourable outcomes)/(Total number of event)
`= 29/2400`

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

Answer: Number of families with income Rs. 16000 or more per month and owning exactly 1 vehicle = 579
Hence, `P(E) = 579/2400`

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

Answer: Number of families earning less than Rs. 7000 per month and does not own any vehicle = 10
Hence, `P(E) = 10/2400 = 1/240`

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

Answer: Number of families earning Rs. 13000 – 16000 per month and owning more than 2 vehicles = 25
Hence, `P(E) = 25/2400 = 1/96`

(v) owning not more than 1 vehicle.

Answer: Number of families owning not more than 1 vehicle
`= 10 + 1 + 2 + 1 + 160 + 305 + 535 + 469 + 579 = 2062`
Hence, `P(E) = 2062/2400 = 1031/1200`



Ex 15.1 Part 1

Ex 15.1 Part 2