﻿ 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. from Mathematics Probability Class 9 Jammu and Kashmir Board # 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.

Let E be the event of hitting the boundary.

Then, ∴ Probability of not hitting the boundary
= 1 – Probability of hitting the boundary
= 1 – P(E) = 1 – 0.2 = 0.8.

1008 Views

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 71 28

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

Total number of times the three coins are tossed = 200
Number of times when 2 heads appear = 72
∴ Probability of 2 heads coming up 1243 Views

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

(0 2 girls (ii) 1 girl (iii) No girl.

Also check whether the sum of these probabilities is 1.

Total number of families
= 475 + 814 + 211 = 1500
(i) Probability of a family, chosen at random,

having 2 girls = (ii) Probability of a family, chosen at random,

having 1 girl (iii) Probability of a family, chosen at random,

having no girl Sum of these probabilities Hence, the sum is checked.

958 Views

In a particular section of Class IX, 40 students were asked about the months of their birth, the following graph was prepared for the data so obtained. Find the probability that a student of the class was bom in August. Total number of students bom in the year =3+4+2+2+5+1+2+6+3+4+4+4=40
Number of students bom in August = 6
∴ Probability that a student of the class was

born in August = 373 Views

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 Vehicles per family (in र) 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  र 10000–13000 per month and owning exactly 2 vehicles.
(ii) earning र 16000 or more per month and owning exactly l vehicle.
(iii) earning less than र 7000 per month and does not own any vehicle.
(iv) earning र 13000–16000 per month and owning more than 2 vehicles.
(v) owning not more than I vehicle.

Total number of families selected = 2400
(i) Number of families earning र10000–13000 per month and owning exactly 2 vehicles = 29
∴ Probability that the family chosen is earning र10000–13000 per month and owning

exactly 2 vehicles = (ii) Number of families earning र16000 or more per month and owning exactly 1 vehicle = 579

∴ Probability that the family chosen is earning र16000 or more per month and owning

exactly 1 vehicle = (iii) Number of families earning less than र 7000 per month and does not own any vehicle =10

Probability that the family chosen is earning less than र7000 per month and does not own any vehicle = (iv) Number of families earning र13000–16000 per month and owning more than 2 vehicles = 25

∴ Probability that the family chosen is earning र13000–16000 per month and owning
more than 2 vehicles = (v) Number of families owning not more than 1 vehicle

= Number of families owning 0 vehicle + Number of families owning 1 vehicle = (10 + 0+ 1 + 2 + 1) +(160 + 305 + 535 + 469 +579) = 14 + 2048 = 2062
∴ Probability that the family chosen owns not more than 1 vehicle = 542 Views