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Math Help - Probability question

  1. #1
    jjc
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    Probability question

    A man writes 5 letters, one each to A, B, C, D and E. Each letter is placed in a separate envelope and sealed. He then addressed the envelopes, at random, one each to A, B, C, D and E.

    Find the probability that both of the letters to A and B are in incorrect envelopes.

    The answer is given as 13/20 but I can't figure out why. Can anyone help?
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  2. #2
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    Quote Originally Posted by jjc View Post
    A man writes 5 letters, one each to A, B, C, D and E. Each letter is placed in a separate envelope and sealed. He then addressed the envelopes, at random, one each to A, B, C, D and E. Find the probability that both of the letters to A and B are in incorrect envelopes. The answer is given as 13/20
    The number of ways that A or B is in the correct envelope is give by:
    P(A \cup B) = P(A) + P(B) - P(A \cap B) = \frac{{2(4!) - 3!}}{{5!}}
    Now, for neither to get the correct letter subtract from 1.
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  3. #3
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    Hello, jjc!

    A man writes 5 letters, one each to A, B, C, D and E.
    Each letter is placed in a separate envelope and sealed.
    He then addressed the envelopes, at random, one each to A, B, C, D and E.

    Find the probability that both of the letters to A and B are in incorrect envelopes.

    Answer: \frac{13}{20}
    Here is a logical (but very primitive) approach . . .


    The letters are: . A,B,C,D,E

    Their envelopes are: . a,b,c,d,e


    There are two cases to consider:
    . . (1) Letter A is in envelope b.
    . . (2) Letter A is not in envelope b.


    (1) Letter A is in envelope b: . P(\text{A in b}) \:=\:\frac{1}{5}

    Then letter B can go in any of the other four envelopes: \{a,c,d,e\}
    . . P(\text{B wrong}) \:=\:\frac{4}{4} \:=\:1

    Hence: . P(\text{A in b }\wedge\text{ B wrong}) \:=\:\frac{1}{5}\cdot1 \:=\:{\color{blue}\frac{1}{5}}


    (2) Letter A is in not in envelope b ... it is in \{c,d,e\}
    . . P(\text{A not in b}) \:=\:\frac{3}{5}

    Then letter B must be in one of the three remaining envelopes.
    . . P(\text{B wrong}) \:=\:\frac{3}{4}

    Hence: . P(\text{A not in b} \wedge\text{ B wrong}) \:=\:\frac{3}{5}\cdot\frac{3}{4} \:=\:{\color{blue}\frac{9}{20}}


    Therefore: . P(\text{A wrong }\wedge\text{ B wrong}) \;=\;\frac{1}{5} + \frac{9}{20} \;=\;{\bf{\color{red}\frac{13}{20}}}

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