Question

A clerk was asked to mail three report cards to three student. He addresses three envelop but unfortunately paid no attention to which report card be put in which envelop. What is the probability that exactly one of the students received his or her own card?

Answer 1

Let $A$ be the first report card in correct envelope.

Let $B$ be the second report card in correct envelope.

Let $C$ be the third report card in correct envelope.

So, $P\left(A\right)=P\left(B\right)=P\left(C\right)=\frac{1}{3}$

Required probability$=P[(A\cap B^{\prime }\cap C^{\prime })\cup (A^{\prime }\cap B\cap C^{\prime })\cup (A^{\prime }\cap B^{\prime }\cap C)]$

$=P[(A\cap B^{\prime }\cap C^{\prime })+(A^{\prime }\cap B\cap C^{\prime })+(A^{\prime }\cap B^{\prime }\cap C)]$

$=P\left(A\right)P\left(\frac{B^{\prime }}{A}\right)P\left(\frac{C^{\prime }}{A\cap B^{\prime }}\right)+P\left(B\right)P\left(\frac{A^{\prime }}{B}\right)P\left(\frac{C^{\prime }}{B\cap A^{\prime }}\right)+P\left(C\right)P\left(\frac{A^{\prime }}{C}\right)P\left(\frac{B^{\prime }}{A\cap A^{\prime }}\right)$

$=(\frac{1}{3}\times \frac{1}{2}\times 1)+(\frac{1}{3}\times \frac{1}{2}\times 1)+(\frac{1}{3}\times \frac{1}{2}\times 1)$

$=3\times \frac{1}{3}\times \frac{1}{2}\times 1=\frac{1}{2}$