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A man has n keys on a key ring, one of which opens the door to his apartment. Having celebrated a bit too much one evening, he returns home only to find himself unable to distinguish one key from another. Resourcesful, he works out a fiendishly clever plan: He will choose a key at random and try it. If it fails to open the door, he will discard it and choose at random one of the remaining n-1 keys, and so on. Clearly, the probability that he gains entrance with the first key he selects is 1/n. What is the probability that the door opens with the third key

Answer :

okpalawalter8

Answer:

The value is   [tex]P(3) = \frac{1}{n}[/tex]

Step-by-step explanation:

From the question we are told that

   The number of keys is  n

    The number of keys remaining after the first key is chosen is  n-1

    The probability that he gains entrance with the first key he selects is [tex]\frac{1}{n}[/tex]

      Generally the probability that the first key does not open the door is

             [tex]p(F_1) = 1 - \frac{1}{n} = \frac{n-1}{n}[/tex]

      Generally the number of keys remaining after the second key is chosen is  

         n-2

      Generally the probability that he gains entrance with the second key he selects is

        [tex]\frac{1}{n-1}[/tex]

       Generally the probability that the second key does not opens the door is

        [tex]P(F_2) = 1- \frac{1}{n-1} = \frac{n-2}{n-1}[/tex]

   Generally the probability that he gains entrance with the third key he selects is

         [tex]\frac{1}{n-2}[/tex]

Generally the probability that the door opens with the third key

        [tex]P(3) = p(F_1) * P(F_2) * \frac{1}{n-2}[/tex]

=>       [tex]P(3) =  \frac{n-1}{n} *   \frac{n-2}{n-1} * \frac{n-2}{n-1}[/tex]

=>        [tex]P(3) =  \frac{1}{n}[/tex]

Note :  

      All the outcome of the events are independent

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