There are four letters and four envelopes, the letters are placed into the envelopes at random, the probability that all letters are placed in the wrong envelopes is

Question

There are four letters and four envelopes, the letters are placed into the envelopes at random, the probability that all letters are placed in the wrong envelopes is (A) (1/8) (B) (3/8) (C) (5/8) (D) 1

Tardigrade Admin · Accepted Answer

We know from the above given formula that probability that no letter is in right envelope out of

n

letters and

n

envelopes is given by

= [\frac{1}{2 !} - \frac{1}{3 !} + \frac{1}{4 !} - \dots\dots\dots . (- 1)^{n} \frac{1}{n !}]

Since all 4 letters are to be placed in wrong envelopes then required probability

= [\frac{1}{2 !} - \frac{1}{3 !} + \frac{1}{4 !}] = \frac{1}{2} - \frac{1}{6} + \frac{1}{24} = \frac{3}{8}

Q. There are four letters and four envelopes, the letters are placed into the envelopes at random, the probability that all letters are placed in the wrong envelopes is

$\frac{1}{8}$

$\frac{3}{8}$

$\frac{5}{8}$

$1$

Q. There are four letters and four envelopes, the letters are placed into the envelopes at random, the probability that all letters are placed in the wrong envelopes is

81​

83​

85​

1

$\frac{1}{8}$

$\frac{3}{8}$

$\frac{5}{8}$

$1$