Question

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

No. of girls in a family	2	1	0
No. of families	475	814	211

Compute the probability of a family, chosen at random, having
(i) 2 girls
(ii) 1 girl
(iii) No girl
Also check whether the sum of these probability is 1.

Answer 1

Hint: To find out the probability of a desired outcome we are provided with the formulae
P(A) =n(E)/n(S)
Where,
P(A) = Probability of an event
n(E) = Number of desired outcome
n(S) = Total number of outcomes


Complete step by step explanation:
Case (i): Family chosen at random have 2 girls
1. Total number of families = 1500
i.e. n(S), which is the total number of outcome = 1500
2. Number of families chosen at random having 2 girls = 475
i.e. n(E), which is the desired outcome = 475
3. P(A)1 is the probability of family chosen at randomly having 2 girls
Using formulae P(A) = n(E)/n(S)
And putting the values of n(E) and n(S) from step 1 and step 2
P(A)1 = \[\dfrac{{475}}{{1500}}\]
P(A)1 Probability of family chosen at random having 2 girls = \[\dfrac{{475}}{{1500}}\]


Case (ii): Family chosen at random have 1 girl
4. Total number of families = 1500
i.e. n(S), which is the total number of outcome = 1500
5. Number of family chosen at random having 1 girls = 814
i.e. n(E), which is the desired outcome = 814
6. P(A)2 is the probability of family chosen at randomly having 1 girls
Using formulae P(A) = n(E)/n(S)
And putting the values of n(E) and n(S) from step 4 and step 5
P(A)2 = $\dfrac{{814}}{{1500}}$
P(A)2 Probability of family chosen at random having 1 girl = $\dfrac{{814}}{{1500}}$
Case (iii): Family chosen at random having 0 girl
7. Total number of families = 1500
i.e. n(S), which is the total number of outcome = 1500
8. Number of family chosen at random having 0 girls = 211
i.e. n(E), which is the desired outcome = 211
9. P(A)3 is the probability of family chosen at randomly having 0 girls
Using formulae P(A) = n(E)/n(S)
And putting the values of n(E) and n(S) from step 7 and step 8
P(A)3 =$\dfrac{{211}}{{1500}}$
P(A)3 Probability of family chosen at random having 0 girls = $\dfrac{{211}}{{1500}}$
10. To check whether the sum of probability of sum is 1 or not
= P(A)1 + P(A)2 + P(A)3
= $\dfrac{{475}}{{1500}} + \dfrac{{814}}{{1500}} + \dfrac{{211}}{{1500}}$
= $\dfrac{{475 + 814 + 211}}{{1500}}$
= $\dfrac{{1500}}{{1500}}$
= 1
Hence proved that the sum of the probabilities of all the three cases is 1.


Note: Always note that Sum of the probabilities of all the cases should always be equal to 1.If the sum of the probabilities is not equal to 1 ,then there would be something wrong in the intermediate steps.This is also a method to verify the result