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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 |
(i) No girl
(ii) 1 girl
(iii) 2 girls
(iv) at most one girl
(v) more girls than boys
Solution
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
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