A couple decides to have \(5\) children. Each child is equally likely to be a boy or a girl. What is the probability that the couple has exactly \(3\) girls and \(2\) boys?
A. \(\dfrac1{32}\)
B. \(\dfrac1{16}\)
C. \(\dfrac5{32}\)
D. \(\dfrac5{16}\)
E. \(\dfrac16\)
Answer: D
Source: eGMAT
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A couple decides to have \(5\) children. Each child is equally likely to be a boy or a girl. What is the probability
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Number of possibilities is \(2,\) it's either a boy or a girl.Gmat_mission wrote: ↑Fri Jan 28, 2022 12:14 pmA couple decides to have \(5\) children. Each child is equally likely to be a boy or a girl. What is the probability that the couple has exactly \(3\) girls and \(2\) boys?
A. \(\dfrac1{32}\)
B. \(\dfrac1{16}\)
C. \(\dfrac5{32}\)
D. \(\dfrac5{16}\)
E. \(\dfrac16\)
Answer: D
Source: eGMAT
It is mentioned that \(3\) girls and \(2\) boys should be the set.
When the firstborn is a \(B: B*B*G*G*G\) (This can be done in \(4C3\) ways by rearranging the \(B\) and \(G\)) \(=\dfrac{1}{2}*\dfrac{1}{2}*\dfrac{1}{2}*\dfrac{1}{2}*\dfrac{1}{2}*4= 4\) ways.
When firstborn is a \(G: G*B*B*G*G\) (This can be done in \(4C2\) ways) \(= \dfrac{1}{2}*\dfrac{1}{2}*\dfrac{1}{2}*\dfrac{1}{2}*\dfrac{1}{2}*6= 6\) ways.
Adding the two scenarios\(: \dfrac{4}{32}+\dfrac{6}{32}=\dfrac{10}{32} =\dfrac{5}{16}\)
Hope this helps!