Question

A man has 3 coins $A , B$ and $C , A$ is unbiased, the probability that a head will result when $B$ is tossed is $2 / 3$, the probability that a head will result when $C$ is tossed is $1 / 3$. One of the coin chosen at random and is tossed 3 times, giving a total of two heads and one tail.

List-I		List-I I
P	The probability that the chosen coin was A	1	32/75
Q	The probability that the chosen coin was B	2	1/2
R	The probability that the fourth toss will result in head	3	209/1296
S	The probability that the fourth toss will result in tail	4	9/25
		5	215/1296

• A. $P \rightarrow 4 ; Q \rightarrow 1 ; R \rightarrow 2 ; S \rightarrow 3$
• B. $P \rightarrow 5 ; Q \rightarrow 4 ; R \rightarrow 2 ; S \rightarrow 3$
• C. $P \rightarrow 3 ; Q \rightarrow 2 ; R \rightarrow 5 ; S \rightarrow 1$
• D. $P \rightarrow 4 ; Q \rightarrow 1 ; R \rightarrow 2 ; S \rightarrow 2$

Answer 1

Answer: (D)

Let $E$ be the event of getting 2 head and 1 tail
$ P(A)=P(B)=P(C)=\frac{1}{3} $
Using Bayes' theorem $P ( A / E )=\frac{9}{25}$,
$ P(B / E)=\frac{32}{75} $