While it is often assumed that the probabilities of having a boy or a girl are the same, the actual probability of having a boy is slightly higher at 0.51. Suppose a couple plans to have 3 children. What is the probability that exactly 2 of them will be boys?
+118723 The 3 at the front accounts for the fact that the girl can be the first or the second or the 3rd child.
If the three was not there you would have the prob that the girl was in a specific position. 1st or second or 3rd.
If found by formula it would be 3C1 one girl out of three (or 3C2 two boys out of 3) they are the same.
$$3C1=\frac{3!}{1!(3-1)!}=\frac{3!}{2!}=3$$
+33663 boy boy girl 0.51*0.51*0.49
boy girl boy 0.51*0.49*0.51
girl boy boy 0.49*0.51*0.51
So probability is
$${\mathtt{p}} = {\mathtt{3}}{\mathtt{\,\times\,}}{{\mathtt{0.51}}}^{{\mathtt{2}}}{\mathtt{\,\times\,}}{\mathtt{0.49}} \Rightarrow {\mathtt{p}} = {\mathtt{0.382\: \!347}}$$
.
+118723 The 3 at the front accounts for the fact that the girl can be the first or the second or the 3rd child.
If the three was not there you would have the prob that the girl was in a specific position. 1st or second or 3rd.
If found by formula it would be 3C1 one girl out of three (or 3C2 two boys out of 3) they are the same.
$$3C1=\frac{3!}{1!(3-1)!}=\frac{3!}{2!}=3$$
