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.

Question

For two events A and B, if \(P(A) = P(A/B) = \frac{1}{4}\) and \(P(B/A) = \frac{1}{2}\), then which of the following is not true?

Answer 1

Given:

Three letters are to be placed into three addressed envelopes.
Each envelope receives exactly one letter.

Letters are inserted at random.

Step 1: Find total number of possible arrangements

Total ways of placing 3 letters into 3 envelopes =

3! = 6

Step 2: Find number of arrangements with no letter in its proper envelope

Such arrangements are called derangements.

Number of derangements of 3 objects =

!3 = 2

(These are: (2,3,1) and (3,1,2))

Step 3: Find number of favourable arrangements

Favourable cases = Total arrangements − Derangements

= 6 − 2

= 4

Step 4: Calculate probability

Probability =

(Number of favourable outcomes) / (Total number of outcomes)

= 4 / 6

= 2 / 3

Final Answer:

The probability that at least one letter is in its proper envelope is
2 / 3

Solution and Explanation