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In a family having 3 children, there may be no girl, one girl, two girls or three girls. So, the probability of each is $\dfrac{1}{4}$ . Is this correct? Justify your answer.

Answer Verified Verified
Hint: Count the total cases properly. Probability is $\dfrac{{{\text{Number of favourable cases}}}}{{{\text{Number of total cases}}}}$ .

No! it’ not correct.
Let boys be B and girls be G. the outcomes can be BBB, GGG, BBG, BGB, GBB, GGB, GBG, BGG then probability of 3 girls $ = \dfrac{1}{8}$. So, the probabilities are as follows:
Probability of 0 girl is $\dfrac{1}{8}$
Probability of 1 girl is $\dfrac{3}{8}$
Probability of 2 girl is $\dfrac{3}{8}$
Probability of 3 girl is $\dfrac{1}{8}$

Note: counting is the most important aspect in order to correct the probability problem. Once you are done with counting and probability is nothing but a ratio.
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