In a family of 3 children find probablyty of having atleast one boy
Answers
the probability is 1 by 3
The answer is 7/8. 7 out of these 8 outcomes have at least one boy {BBB,BBG,BGG,BGB,GBB,GBG,GGB,GGG}.
The important question is why did we chose this set as our basis and why is the probability of each of these equal.
We might have as well said that the events were { 3 boys, 1 girl 2 boys, 2 girls 1 boy, 3 girls } and said that the probabilities of each of these are equal.
However that calculation would be wrong since it would require us to make assumptions that would contradict with the facts of the the problem.
While solving probability problems we should avoid making unfounded assumptions if it is unnecessary. In this problem we know the following facts from our knowledge about the statistics of sex determination in children.
The probability of having a boy is equal to that of getting a girl.
The outcome of one birth is independent of that of the other.
In terms of probabilities this can be written as:
P(B)=P(G)
P(BB)/P(GB)=P(BG)/P(GG)
Where P(B) means probability of having a boy in the case of a single birth and P(GB) means the probability of having a girl followed by a boy in two births.
Doing some algebra on the above equations we can reach:
P(BBB)=P(BBG)=P(BGG)=... and so on.
And using the fact that the sum of probabilities of all individual exclusive events is 1, we can calculate that each of these probabilities is 1/8. Given that 7 of these cases have at least one boy, the total probability for at least one boy becomes 7/8.