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Junior | Next Rank: 30 Posts
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by Mumbai » Fri Jul 01, 2011 2:05 am
How to solve this?

Tanya prepared 4 different letters to be sent to 4 different addresses. For each letter she prepared an envelope with 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

[spoiler]Answer - d - 1/3[/spoiler]
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by Frankenstein » Fri Jul 01, 2011 2:33 am
Hi,
Let the 4 letters be a1,a2,a3,a4 and their corresponding envelopes be e1,e2,e3,e4 respectively
Putting a1 in e1 and the others in unmatched envelopes is :
e1 e2 e3 e4
1)a1 a3 a4 a2
2)a1 a4 a2 a3 --> 2ways
Similarly each of a1,a2,a3,a4 can be placed in its corresponding envelopes in 2*4 = 8 ways
Total number of ways on putting 4 letters in 4 envelopes is 4! = 24
So, probability is 8/24 = 1/3
Cheers!

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by testprepDublin » Fri Jul 01, 2011 5:12 am
There are four different ways that only one letter will be put in the right envelope. The odds of the one letter being put in the correct envelope are worked out below:

The probability of the letter 1 being correct = 1/4, the probability of letter 2 being incorrect = 2/3, the probability of letters 3 and 4 being incorrect = 1/2. So 1/4*2/3*1/2 = 1/12.

Since there are 4 different letters (and 4 different ways that only 1 letter will be put into the correct envelope), we get (1/12)*4 = 1/3. The answer is D.
Deirdre at testprepdublin.com
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by GMATGuruNY » Fri Jul 01, 2011 9:21 am
Mumbai wrote:How to solve this?

Tanya prepared 4 different letters to be sent to 4 different addresses. For each letter she prepared an envelope with 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

[spoiler]Answer - d - 1/3[/spoiler]
I posted a solution here:

https://www.beatthegmat.com/difficult-pr ... tml#344738

The post above also describes how to determine the probability that NO envelope is given the correct letter (an even trickier problem).
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by Mumbai » Fri Jul 08, 2011 3:57 am
thank you guys... this helps!!
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