One Right three wrong.

P = P (Getting one in correct envelope) AND
P (Getting second in wrong envelope) AND
P (Getting 3rd in wrong envelope) AND
P (Getting 4th in wrong envelope)

= (1/4) * (2/3) * (1/2) * 1
= 1/12

So there are 4 such cases RWWW OR WRWW OR WWRW OR WWWR

Thus, 4*(1/12) = 1/3

IMOD

ugoyal wrote:Tanya prepared 4 different letters to be sent to 4 different addresses. For each letter, she prepared an envelope with its correct address. If the 4 letters are to be put into the 4 envelopes at random, what is the probability that only 1 letter will be put into the envelope with its correct address?

A. 1/24

B. 1/8

C. 1/4

D. 1/3

E. 3/8

Let's call the envelopes E1, E2, E3 and E4.

P(only E1 gets the correct letter):

P(E1 gets the correct letter) = 1/4 (4 letters total, 1 of them correct)
P(E2 gets the wrong letter) = 2/3 (3 letters left, 2 of them wrong)
P(E3 gets the wrong letter) = 1/2 (2 letters left, 1 of them wrong)
P(E4 gets the wrong letter) = 1/1 (1 letter left, and it must be wrong since we placed the correct letter in either E2 or E3)

Since we need all of these events to happen, we multiply the fractions:

1/4 * 2/3 * 1/2 * 1/1 = 1/12.

Since each envelope has the same probability of getting the correct letter and we have 4 envelopes total, we need to multiply by 4:

4 * 1/12 = 1/3.