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Prep Questions...please help me

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Prep Questions...please help me

by ankit1383 » Sun Sep 28, 2008 9:18 am
Tanya prepared 4 different letters to be sent to 4 different addresses.For each letter,she prepared an envelope with its correct address.If 4 letters are to be put in to 4 envelope at random,what is the probability that only 1 letter will be put in the address with the correct address?

1 1/24
2 1/8
3 1/4
4 1/3
5 3/8
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by anniev2 » Sun Sep 28, 2008 9:22 am
is the answer 1/8?
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Answer

by ankit1383 » Sun Sep 28, 2008 9:28 am
1/3.....................please explain the answer.............
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by Morgoth » Sun Sep 28, 2008 9:55 am
4 envelops, 4 letters

1 right, 3 wrong.

probability first one goes into correct envelop = 1/4
probability second one goes into incorrect envelop = 1-1/3 = 2/3
probability third one goes into incorrect envelop = 1-1/2 = 1/2
the fourth one will automatically will go into incorrect envelop = 1

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


Now we have four cases i.e second going into correct, then third one going into correct envelope and lastly fourth one going into correct one.

Thus,

total probability = 1/12 + 1/12 + 1/12 + 1/12 = 4/12 = 1/3

Hope this helps
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