Probability

Roland2rule wrote:Tanya prepared 4 different letters to 4 different addresses. For each letter, she prepared one envelope with its correct address. If the 4 letters are to be put into the four envelopes at random, what is the probability that only one 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

OA is D

Do I need to use the T-tree to solve this question? pls, I need an expert thought.

The successful letter has 1/4 chance of being placed in its envelope.

There are now 3 envelopes remaining. There is a 2/3 chance the next letter placed will be in the incorrect envelope.

There are now 2 envelopes remaining. There is a 1/2 chance the next letter will be in the incorrect envelope.

There is now 1 envelope remaining and there is a 100% chance that the wrong letter will be inserted.

Multiplying these out: 1/4*2/3*1/2*1 = 2/24. Multiply this by 4 to reflect that you could have picked any one of the 4 letters to be placed in its proper envelope = 8/24 = [spoiler]1/3, D[/spoiler]