Question

If birth to a male child and birth to a female child are equally probable, then what is the probability that at least one of the three children born to a couple is male?
(a) \[\dfrac{4}{5}\]
(b) \[\dfrac{7}{8}\]
(c) \[\dfrac{8}{7}\]
(d) \[\dfrac{1}{2}\]

Answer 1

Hint: First of all assume the probability of a male child is x. The probability of a female child will also be x because the probability of birth of both males and females are equally probable. The probability of birth of a female child will be x. We know that the total probability of a male child and a female child is 1. Using this we get the probability of birth of a male and a female child. Now, we have to find the probability of at least one male child. It means we don’t have any restrictions on the maximum number of children. First, find the probability of exactly one male child. Then, find the probability of exactly two male children. Then, finally, get the probability of exactly three male children. Now, add these probabilities and solve it further.

Complete step-by-step solution -
According to the question, it is given that the birth of a male child and a female child are equally probable.
Let the probability of birth of a male child be x. Since the probability of the birth of a male child and a female child are equally probable, so the probability of birth of a female child is also x.
We know that the total probability of a male child and a female child is 1.
So, Total Probability = Probability of birth of a male child + Probability of birth of a male child
\[\begin{align}
  & 1=x+x \\
 & \Rightarrow 1=2x \\
 & \Rightarrow \dfrac{1}{2}=x \\
\end{align}\]
Now, we have the probability of birth of a male child and a female child is \[\dfrac{1}{2}\] .
We have to find the probability that at least one of the three children born to a couple is male.
The probability that the \[{{1}^{st}}\] child is male and rest are female children = \[\left( \dfrac{1}{2} \right){{\left( \dfrac{1}{2} \right)}^{2}}=\dfrac{1}{8}\] .
The probability that the \[{{2}^{nd}}\] child is male and rest are female children = \[\left( \dfrac{1}{2} \right)\left( \dfrac{1}{2} \right)\left( \dfrac{1}{2} \right)=\dfrac{1}{8}\] .
The probability that the \[{{3}^{rd}}\] child is male and rest are female children = \[{{\left( \dfrac{1}{2} \right)}^{2}}\left( \dfrac{1}{2} \right)=\dfrac{1}{8}\] .
Probability of exactly one child is male and rest are female child = \[\dfrac{1}{8}+\dfrac{1}{8}+\dfrac{1}{8}=\dfrac{3}{8}\] ……………….(1)
The probability that the first and second child is male and rest is female= \[{{\left( \dfrac{1}{2} \right)}^{2}}\left( \dfrac{1}{2} \right)=\dfrac{1}{8}\] .
The probability that the second and third child is male and first is female= \[\left( \dfrac{1}{2} \right){{\left( \dfrac{1}{2} \right)}^{2}}=\dfrac{1}{8}\] .
The probability that the first and the third child are male and second is female = \[\left( \dfrac{1}{2} \right)\left( \dfrac{1}{2} \right)\left( \dfrac{1}{2} \right)=\dfrac{1}{8}\] .
Probability of exactly two male children = \[\dfrac{1}{8}+\dfrac{1}{8}+\dfrac{1}{8}=\dfrac{3}{8}\] ………………..(2)
The probability that all the child is male = \[{{\left( \dfrac{1}{2} \right)}^{3}}=\dfrac{1}{8}\] ………………….(3)
Probability of at least one child = Probability of exactly one child is male and rest are female child + Probability of exactly two male children + Probability that all the child is male ………………….(4)
From equation (1), equation (2), equation (3) and equation (4), we have
Probability of at least one child = \[\dfrac{3}{8}+\dfrac{3}{8}+\dfrac{1}{8}=\dfrac{7}{8}\] .
Therefore, the probability of at least one male child is \[\dfrac{7}{8}\] .
Hence, the correct option is (B).

Note: We can also solve this question using another method. In the question, it is asked the probability of at least one male child. The maximum number of male children can be 3. If we deduct the probability of no male child from the total probability then we will get the probability of at least one male child.
The probability that all no children are male is equal to the probability that all the children are female.
Probability of all-female child = \[{{\left( \dfrac{1}{2} \right)}^{3}}=\dfrac{1}{8}\] .
The probability that at least one child is male = \[1-\dfrac{1}{8}=\dfrac{7}{8}\] .
Therefore, the probability of at least one male child is \[\dfrac{7}{8}\] .