A card from a pasck of 52 card is lost.From the remaining cards of the pack,two cards are drawn and are found to be both diamonds.Find the probability of the lost card being a diamond.
Solution and Explanation
Let E1=the missing card is a diamond,E2=the missing card is a spade,E3=the missing card is a club,E4=the missing card is a heart and A=drawing of two heart cards from the remaining cards.
Now P(E1)=\(\frac{13}{52}\)=\(\frac{1}{4}\),P(E2)=\(\frac{13}{52}\)=\(\frac{1}{4}\),P(E3)=\(\frac{13}{52}\)=\(\frac{1}{4}\),P(E4)=\(\frac{13}{52}\)=\(\frac{1}{4}\)
P(A|E1)=P(drawing 2 heart cards given that one diamond card is missing)=\(\frac{C(12,2)}{C(51,2)}\)
By Bayes' theorem,
\(P(E_1|A)=\frac{P(E_1)P(A|E_1)}{P(E_1)P(A|E_1)+P(E_2)P(A|E_2)+P(E_3)P(A|E_3)+P(E_4)P(A|E_4)}\)
\(=\frac{1}{4}\times \frac{C(12,2)}{C(51,2)}\)
=\(\frac{66}{66}\)+\(78+78+7\)\(8\)=\(\frac{11}{50}\)
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Concepts Used:
Bayes Theorem
Bayes’ Theorem is a part of the conditional probability that helps in finding the probability of an event, based on previous knowledge of conditions that might be related to that event.
Mathematically, Bayes’ Theorem is stated as:-
\(P(A|B)=\frac{P(B|A)P(A)}{P(B)}\)
where,
- Events A and B are mutually exhaustive events.
- P(A) and P(B) are the probabilities of events A and B, respectively.
- P(A|B) is the conditional probability of the happening of event A, given that event B has happened.
- P(B|A) is the conditional probability of the happening of event B, given that event A has already happened.
This formula confines well as long as there are only two events. However, Bayes’ Theorem is not confined to two events. Hence, for more events.