Question

Find the probability that the first child of a family with five children is a boy or that the last two children of the family are girls, for the condition that the probability of boy is 0.51 ?

Comments

can you have the answer of this question  ?

Answer 1

This is my approach .

the probability of 1 st child is boy 51/100.

the probability of last two child is girl (49/100)² .

the probability of 1 st child is boy and the probability of last two child is girl (51*49*49/100³) .

so the probability of 1 st child is boy or the probability of last two child is girl = (51/100) +  (49/100)² - (51*49*49/100³)

= 0.6276 

Answer 2

Another approach, since the probability of boy and girl are almost same, let us consider both of them to be 0.5. Now there are 5 children and each of them could be a boy or a girl so using product rule there are total 2^5=32 possible outcomes. Out of these 32, if it is fixed that first baby is a boy then there are 2^4 = 16 possibilities for remaining 4 babies, and Similarly if it is fixed that last two babies are girls then there are 2^3 = 8 possibilities for remaining 3 babies. Suppose a case when first baby is a boy and last two are girls, then there are 2^2 = 4 possibilities. These 4 possibilities have been counted twice so we have to subtract it once. Thus total favourable outcomes = 16+8-4=20, And total possible outcomes = 32. So probability = 20/32 = 0.625. So answer will be around 0.625. It can be done without calculator.

Comments

yes, firstly i tried it with this approach but the probabilities of boy and girl are not same. It can easily be solved by the approach that pranay have given.

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Yeah but there is a trade off between time and accuracy, this approach will require less time and calculation by costing you accuracy. However it is accurate upto one decimal places and could be quite useful in multiple choice questions.

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yup!