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despoto
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Anna needs to send 4 letters to 4 different addresses. She 22 Feb 2006, 15:42
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Anna needs to send 4 letters to 4 different addresses. She puts the addresses on the envelopes and then randomly puts the letters in each of the four envelopes. What is the probability that only one letter will be sent to the right address?

Please explain your reasoning. Thank you.



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kook44
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Anna needs to send 4 letters to 4 different addresses. She 22 Feb 2006, 16:03
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is it
1/4 * 2/3 * 1/2 = 1/12 ?
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despoto
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Anna needs to send 4 letters to 4 different addresses. She 22 Feb 2006, 16:11
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No, the right answer is 1/3 but I don't know why... If someone can help, thank you.
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ps_dahiya
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Anna needs to send 4 letters to 4 different addresses. She 22 Feb 2006, 18:22
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L - Letter
E - Envelope

L1-L2-L3-L4
E1-E2-E3-E4

Total number of cases = 4! = 24

Cases when only one latter is sent to right destination. Lesta say only L1 is sent to right destination.

L1-L2-L3-L4

E1-E3-E4-E2
E1-E4-E2-E3

There are only two cases when only E1 is sent to right destination.
Same way for E2, E3 and E4. So tatal cases = 2 * 4 = 8

Prob = 8/24 = 1/3
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conocieur
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Anna needs to send 4 letters to 4 different addresses. She 22 Feb 2006, 18:25
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kook44
is it
1/4 * 2/3 * 1/2 = 1/12 ?
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you were close kook, the only thing is that it can be done in 4c1 ways

so 1/12 * 4 is 1/3
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kook44
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Anna needs to send 4 letters to 4 different addresses. She 22 Feb 2006, 20:42
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ok. i was trying to multpily the probability of four dependent events happening together. (first letter is right AND second is wrong AND third is wrong AND fourth is wrong)

are there any rules of thumb to decide between multipling dependant events and dividing inclusive or exclusive combinations by the total amount of combinations?



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