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 family | 2 | 1 | 0 |
|---|---|---|---|
| Number of families | 475 | 814 | 211 |
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:
| Outcome | 3 heads | 2 heads | 1 head | No head |
|---|---|---|---|---|
| Frequency | 23 | 72 | 77 | 28 |
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 family | 0 | 1 | 2 | Above 2 |
|---|---|---|---|---|
| Less than 7000 | 10 | 160 | 25 | 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.
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`