Results 1 to 3 of 3

Thread: Probability question

  1. #1
    jjc
    jjc is offline
    Newbie
    Joined
    Feb 2008
    Posts
    4

    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?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Aug 2006
    Posts
    21,782
    Thanks
    2824
    Awards
    1
    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:
    $\displaystyle P(A \cup B) = P(A) + P(B) - P(A \cap B) = \frac{{2(4!) - 3!}}{{5!}}$
    Now, for neither to get the correct letter subtract from 1.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    12,028
    Thanks
    848
    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: $\displaystyle \frac{13}{20}$
    Here is a logical (but very primitive) approach . . .


    The letters are: .$\displaystyle A,B,C,D,E$

    Their envelopes are: .$\displaystyle a,b,c,d,e$


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


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

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

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


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

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

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


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

    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 5
    Last Post: Feb 10th 2013, 02:11 PM
  2. Probability question involving (A given B type question )
    Posted in the Advanced Statistics Forum
    Replies: 0
    Last Post: Nov 9th 2009, 09:08 AM
  3. New Probability question
    Posted in the Statistics Forum
    Replies: 2
    Last Post: Jul 15th 2009, 06:50 AM
  4. probability question
    Posted in the Advanced Statistics Forum
    Replies: 4
    Last Post: May 16th 2009, 06:36 AM
  5. A probability question and an Expectations question
    Posted in the Advanced Statistics Forum
    Replies: 1
    Last Post: Jan 29th 2006, 06:13 PM

Search Tags


/mathhelpforum @mathhelpforum