Question

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

Number of girls in a family	0	1	2
Number of families	211	814	475

 

If a family is chosen at random, compute the probability that it has:
(i) No girl
(ii) 1 girl
(iii) 2 girls
(iv) at most one girl
(v) more girls than boys

Answer 1

The total number of trials is 1500.

Remember the empirical or experimental or observed frequency approach to probability.

If n be the total number of trials of an experiment and A is an event associated to it such that A happens in m-trials. Then the empirical probability of happening of event A is denoted by p(A) and is given by

P(A) = `m/n`

(i) Let A be the event of having no girl.

The number of times A happens is 211.

Therefore, we have

P(A) = `211/1500`

=0.1406

(ii) Let B be the event of having one girl.

The number of times B happens is 814.

Therefore, we have

P(B) = `814/1500`

=0.5426

(iii) Let C be the event of having two girls.

The number of times C happens is 475.

Therefore, we have

P(C) = `475/1500`

=0.3166

(iv) Let D be the event of having at most one girl.

The number of times D happens is 211+814=1025.

Therefore, we have

P(D) = `1025/1500`

=0.6833

(v) Let E be the event of having more girls than boys.

The number of times E happens is 475.

Therefore, we have

P(E) = `475/1500`

=0.3166