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(D) The total number of ways to place $3$ distinct letters into $3$ distinct envelopes is given by the permutation of $3$ items,which is $3! = 3 	ime

Vedclass · Accepted Answer

$\frac{1}{6}$

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(D) The total number of ways to place $3$ distinct letters into $3$ distinct envelopes is given by the permutation of $3$ items,which is $3! = 3 	imes 2 	imes 1 = 6$.Out of these $6$ possible arrangements,only $1$ arrangement corresponds to the case where every letter is placed in its correct envelope.Therefore,the required probability is the ratio of the number of favorable outcomes to the total number of possible outcomes:Probability $= \frac{	ext{Number of favorable outcomes}}{	ext{Total number of outcomes}} = \frac{1}{3!} = \frac{1}{6}$.

Three letters are to be sent to different persons,and addresses on the three envelopes are also written. Without looking at the addresses,the probability that all the letters go into the right envelope is equal to

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