Question

Assume that the birth of a boy or girl to a couple to be equally likely, mutually exclusive exhaustive and independent of the other children in the family for a couple having 6 children the probability that their ‘three oldest are boy’ is
\[\begin{align}
  & \text{(A) }\dfrac{20}{64} \\
 & \text{(B) }\dfrac{1}{64} \\
 & \text{(C) }\dfrac{2}{64} \\
 & \text{(D) }\dfrac{8}{64} \\
\end{align}\]

Answer 1

Hint: We know that If two events have an equal probability for occurrence, then these events are said to be equally likely events. If two events cannot occur at the same time then the two events are said to be mutually exclusive and exhaustive events. If two events A and B are independent, then \[P(A\cap B)=P(A)P(B)\]. Let us assume the 6 events as A, B, C, D, E, F. We know that if two events A and B are independent, then \[P(A\cap B)=P(A)P(B)\]. Now we have to calculate the value of \[P(A\cap B\cap C\cap D\cap E\cap F)\]. From this concept, we should find the value of \[P(A\cap B\cap C\cap D\cap E\cap F)\]. This will gives the required probability.

Complete step by step answer:
Before solving the question, we should know that
If two events have an equal probability for occurrence, then these events are said to be equally likely events.
If two events cannot occur at the same time then the two events are said to be mutually exclusive and exhaustive events.
If two events A and B are independent, then \[P(A\cap B)=P(A)P(B)\].
From the question, it is given that the events are equally likely, mutually exclusive, exhaustive and independent.
So, we can say that the probability that the child is boy is equal to the probability that the child is girl.
Let us assume the probability that the child is boy is equal to P(Boy) and the probability for the child to be girl is equal to P(Girl).
\[P(Boy)=P(Girl).....(1)\]
We know that the sum of probabilities of occurrence if all events is equal to 1.
So, we can say that.
\[P(Boy)+P(Girl)=1....(2)\]
Now we will substitute equation (1) in equation (2), then we get
\[\begin{align}
  & \Rightarrow P(Boy)+P(Boy)=1 \\
 & \Rightarrow 2P(Boy)=1 \\
 & \Rightarrow P(Boy)=\dfrac{1}{2}.....(3) \\
\end{align}\]
Now we will substitute equation (3) in equation (1), then we get
\[P(Girl)=\dfrac{1}{2}.....(4)\]
As the events are mutually exclusive and exhaustive, so the occurrence of having a child as a boy and occurrence of having a child as a girl cannot occur together.
From the question, it was given that the 6 children are born in which the three oldest are boys.
Let us assume the 6 events as A, B, C, D, E, F.
We know that if two events A and B are independent, then \[P(A\cap B)=P(A)P(B)\].
Now we have to calculate the value of \[P(A\cap B\cap C\cap D\cap E\cap F)\].
Now we get
\[P(A\cap B\cap C\cap D\cap E\cap F)=P(A)P(B)P(C)P(D)P(E)P(F).....(5)\]
As the first three children are boys, so event A, event B, and event C represent the event that the children born are boys.
So, we get the value of
\[\begin{align}
  & P(A)=\dfrac{1}{2}......(6) \\
 & P(B)=\dfrac{1}{2}.......(7) \\
 & P(C)=\dfrac{1}{2}.......(8) \\
\end{align}\]
Now we will substitute equation (6), equation (7), and equation (8) in equation (5), then we get
\[P(A\cap B\cap C\cap D\cap E\cap F)=\dfrac{1}{2}.\dfrac{1}{2}.\dfrac{1}{2}.P(D).P(E).P(F).....(9)\]
Now we have to find the value of \[P(D).P(E).P(F)\].
We can say that event D, event E and event F may represent the event that the child may be boy or a girl.
Let us write all the assumptions in the form of a table.

Event D	Event E	Event F	P(D)	P(E)	P(F)	Value of \[P(D)P(E)P(F)\]
Boy	Boy	Boy	\[\dfrac{1}{2}\]	\[\dfrac{1}{2}\]	\[\dfrac{1}{2}\]	\[\dfrac{1}{2}.\dfrac{1}{2}.\dfrac{1}{2}=\dfrac{1}{8}\]
Boy	Boy	Girl	\[\dfrac{1}{2}\]	\[\dfrac{1}{2}\]	\[\dfrac{1}{2}\]	\[\dfrac{1}{2}.\dfrac{1}{2}.\dfrac{1}{2}=\dfrac{1}{8}\]
Boy	Girl	Boy	\[\dfrac{1}{2}\]	\[\dfrac{1}{2}\]	\[\dfrac{1}{2}\]	\[\dfrac{1}{2}.\dfrac{1}{2}.\dfrac{1}{2}=\dfrac{1}{8}\]
Girl	Boy	Boy	\[\dfrac{1}{2}\]	\[\dfrac{1}{2}\]	\[\dfrac{1}{2}\]	\[\dfrac{1}{2}.\dfrac{1}{2}.\dfrac{1}{2}=\dfrac{1}{8}\]
Girl	Girl	Boy	\[\dfrac{1}{2}\]	\[\dfrac{1}{2}\]	\[\dfrac{1}{2}\]	\[\dfrac{1}{2}.\dfrac{1}{2}.\dfrac{1}{2}=\dfrac{1}{8}\]
Girl	Boy	Girl	\[\dfrac{1}{2}\]	\[\dfrac{1}{2}\]	\[\dfrac{1}{2}\]	\[\dfrac{1}{2}.\dfrac{1}{2}.\dfrac{1}{2}=\dfrac{1}{8}\]
Girl	Girl	Girl	\[\dfrac{1}{2}\]	\[\dfrac{1}{2}\]	\[\dfrac{1}{2}\]	\[\dfrac{1}{2}.\dfrac{1}{2}.\dfrac{1}{2}=\dfrac{1}{8}\]

From the table, it is clear that all the sum of probabilities of all the events gives us the probability of \[P(D).P(E).P(F)\].
\[P(D).P(E).P(F)=\dfrac{1}{8}+\dfrac{1}{8}+\dfrac{1}{8}+\dfrac{1}{8}+\dfrac{1}{8}+\dfrac{1}{8}+\dfrac{1}{8}+\dfrac{1}{8}=\dfrac{8}{8}......(10)\]
Now let us substitute equation (10) in equation (9), then we get
\[P(A\cap B\cap C\cap D\cap E\cap F)=\dfrac{1}{2}.\dfrac{1}{2}.\dfrac{1}{2}.\dfrac{8}{8}=\dfrac{8}{64}.....(11)\]
From equation (11), it is clear that children the probability that their ‘three oldest are boy’ is equal to \[\dfrac{8}{64}\].
So, option D is correct.

Note:
Students should write each and every possibility in the table. If any event is missed, we will get the wrong value of \[P(D).P(E).P(F)\]. If this value is wrong, then the final probability obtained will also be wrong. Students should write every time in a clear view that no event got missed. Students should also be careful in the calculation part. If a small mistake is done, then we cannot get the correct answer. So, students should do the calculation part in a perfect manner.