Assume that each child born is equally likely to be a boy or a girl. If a family has two children,what is the conditional probability that both are girls given that
Let first and second girls be denoted by G1 and G2 and boys B1 and B2.
∴S = {(G1, G2), (G1, B2), (G2,B1), (B1, B2)}
Let,
A = Both the children are girls = (G1, G2)
B = Youngest child is girl = {(G1, G2), (B1, G2)}
C = at least one is a girl = {(G1, B2),(G1, G2),(B1, G2)}
A∩B = (G1, G2)
⇒P(A∩B) = \(\frac 14\)
A∩C = (G1, G2)
⇒P(A∩C) = \(\frac 14\)
P(B)=\(\frac 24\) and P(C)=\(\frac 34\)
\((i)\) \(P(A|B) = \frac {P(A∩B)}{P(B)}\)
\(P(A|B)=\frac {1/4}{2/4}\)
\(P(A|B)=\frac 12\)
\((ii)\) \(P(A|C)=\frac {P(A∩C)}{P(C)}\)
\(P(A|C)=\frac {1/4}{3/4}\)
\(P(A|C)=\frac 13\)
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Conditional Probability is defined as the occurrence of any event which determines the probability of happening of the other events. Let us imagine a situation, a company allows two days’ holidays in a week apart from Sunday. If Saturday is considered as a holiday, then what would be the probability of Tuesday being considered a holiday as well? To find this out, we use the term Conditional Probability.
P(S | B) = P(B | B) = 1.
Proof of the same: P(S | B) = P(S ∩ B) ⁄ P(B) = P(B) ⁄ P(B) = 1.
[S ∩ B indicates the outcomes common in S and B equals the outcomes in B].
P(B | A), P(A) >0 or, P(A ∩ B) = P(B).P(A | B), P(B) > 0.
This theorem is named as the Multiplication Theorem of Probability.
Proof of the same: As we all know that P(B | A) = P(B ∩ A) / P(A), P(A) ≠ 0.
We can also say that P(B|A) = P(A ∩ B) ⁄ P(A) (as A ∩ B = B ∩ A).
So, P(A ∩ B) = P(A). P(B | A).
Similarly, P(A ∩ B) = P(B). P(A | B).
The interesting information regarding the Multiplication Theorem is that it can further be extended to more than two events and not just limited to the two events. So, one can also use this theorem to find out the conditional probability in terms of A, B, or C.
Read More: Types of Sets
Sometimes students get confused between Conditional Probability and Joint Probability. It is essential to know the differences between the two.