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

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 $P(A/E)=\frac{9}{25}$,
$P(B/E)=\frac{32}{75}$