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

$\begin{array}{1 1}(A)\;\large\frac{2}{3}\\(B)\;\large\frac{5}{3}\\(C)\;\large\frac{7}{3}\\(D)\;\text{None of these}\end{array} $

1 Answer

  • Required probability =$\large\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}}=\frac{n(E)}{n(S)}$
Step 1:
Let the envelope be denoted by $E_1,E_2,E_3$ and the corresponding letters by $L_1,L_2,L_3$
Atleast one letter should be in right envelope.
Let us consider all the favorable outcomes
Step 2:
(i) 1 letter in correct envelope and 2 in wrong envelope.
(ie) $(E_1L_1,E_2L_3,E_3L_2),(E_1L_3,E_2L_2,E_3L_1),(E_1L_2,E_2L_1,E_3L_3)$
(ii) Two letter in correct envelope.
(ie) $(E_1L_1,E_2L_2,E_3L_3)$
$\therefore$ No of favorable outcomes =4
Step 3:
Total no of outcomes =$3!=3\times 2\times 1=6$
$\therefore$ Required probability =$\large\frac{n(E)}{n(S)}$
$\Rightarrow \large\frac{4}{6}$
$\Rightarrow \large\frac{2}{3}$
Hence (A) is the correct answer.
answered Jul 2, 2014 by sreemathi.v

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