Question

Three letters are dictated to three persons and an envelope addressed to each of them, the letters are insearted into the envelopes at randomly so that each envelope contains exactly one letter . Find the probability that atleast one letter is in its proper enbelop .

• A. $\frac{1}{3}$
• B. $\frac{2}{3}$
• C. $\frac{1}{5}$
• D. None of these

Answer 1

The correct option is B $\frac{2}{3}$

Let all three envelopes are denoted by $E_1,E_2$ and $E_3$ and the corresponding letters denoted by $L_1,L_2$ and $L_3$.

$1$ letter in correct envelope and $2$ letters in wrong envelops may be put as

$(E_1{L}_1,E_2{L}_2,E_3{L}_3),(E_2{L}_1,E_2{L}_2,E_2{L}_3),(E_3{L}_1,E_3{L}_2,E_3{L}_3)$

Consequently all three letters are in correct envelope may be put in one way.

Eg: $(E_1{L}_1,E_2{L}_2,E_3{L}_3)$

Hence, number of favourable cases $n\left(E\right)=3+1=4$ ways

Total number of cases $n\left(s\right)=3!=3\times 2=6$ ways

So, $P\left(E\right)=\frac{n\left(E\right)}{n\left(s\right)}=\frac{4}{6}=\frac{2}{3}$.