Geometric or binomial distribution?A monkey is sitting at a simplified keyboard that only includes the keys "a", "b", and "c". The monkey presses the keys at random. Let X be the number of keys pressed until the money has passed all the different keys at least once. For example, if the monkey typed "accaacbcaaac.." then X would equal 7 whereas if the money typed "cbaccaabbcab.." then X would equal 3.a.) What is the probability X ≥ 10?b.) Prove that for an random variable Z taking values in the range {1,2,3,...}, E ( Z ) = Summation from i = 1 to infinity of P ( Z ≥ i ).c.) What's the expected value of X?First, is this a binomial distribution or a geometric distribution? I believe it is a binomial but my other friends says that it is geometric. As for the questions above, for a can I just do 1 − P ( X = 9 ) or 1 − P ( X &lt; 9 ), but I don't know how I will calculate X &lt; 9, I would know how to calculate P ( X = 9 ), I don't know how to do b or c.

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Geometric or binomial distribution?A monkey is sitting at a simplified keyboard that only includes the keys &quot;a&quot;, &quot;b&quot;, and &quot;c&quot;. The monkey presses the keys at random. Let X be the number of keys pressed until the money has passed all the different keys at least once. For example, if the monkey typed &quot;accaacbcaaac..&quot; then X would equal 7 whereas if the money typed &quot;cbaccaabbcab..&quot; then X would equal 3.a.) What is the probability   X  ≥  10?b.) Prove that for an random variable Z taking values in the range {1,2,3,...},   E  (  Z  )  = Summation from   i  =  1 to infinity of   P  (  Z  ≥  i  ).c.) What&#039;s the expected value of X?First, is this a binomial distribution or a geometric distribution? I believe it is a binomial but my other friends says that it is geometric. As for the questions above, for a can I just do   1  −  P  (  X  =  9  )     or     1  −  P  (  X  &amp;lt;  9  ), but I don&#039;t know how I will calculate   X  &amp;lt;  9, I would know how to calculate   P  (  X  =  9  ), I don&#039;t know how to do b or c.

Ismailatirf · Accepted Answer

Step 1We solve only the expectation part, in order to introduce an idea. But to make what the monkey types more interesting, Let us assume that the monkey has 5 letters available.Let       X    1   be the waiting time (the number of key presses) until the first &quot;new&quot; letter. Of course   X  =  1.Let       X    2   be the waiting time between the first new letter, and the second. Let       X    3   be the waiting time between the second new letter and the third. Define       X    4   and       X    5   similarly.Then the total waiting time W is given by   W  =      X    1    +      X    2    +      X    3    +      X    4    +      X    5  . By the linearity of expectation we have   E  (  W  )  =  E  (      X    1    )  +  E  (      X    2    )  +  ⋯  +  E  (      X    5    )  .Step 2Clearly   E  (      X    1    )  =  1.Once we have 1 letter, the probability that a key press produces a new letter is       4    5  . So by a standard result about the geometric distribution,   E  (      X    2    )  =      5    4  .Once we have obtained 2 letters, the probability that a letter is new is       3    5  . Thus   E  (      X    3    )  =      5    3  .Similarly,   E  (      X    4    )  =      5    2   and   E  (      X    5    )  =      5    1  .Add up. To make things look nicer, we bring out a common factor of 5, and reverse the order of summation. We get   E  (  W  )  =  5      (    1    +          1      2        +          1      3        +          1      4        +          1      5        )    .

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