Probability that at least 4 out of 365 times I try, I need more than 5 flips before getting a tailsSuppose I flip a coin until I get a tails. Let X be the number of flips this takes. What is the probability that there are at least 4 days in a year where I needed more than 5 flips to get a tails?Is this kind of question combining geometric and exponential distributions with some kind of conditional probability? I'm unsure how to compute this.

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Probability that at least 4 out of 365 times I try, I need more than 5 flips before getting a tailsSuppose I flip a coin until I get a tails. Let X be the number of flips this takes. What is the probability that there are at least 4 days in a year where I needed more than 5 flips to get a tails?Is this kind of question combining geometric and exponential distributions with some kind of conditional probability? I&#039;m unsure how to compute this.

mentest91k99 · Accepted Answer

Step 1Let       A    i  i be an indicator random variable:      A    i    =      {                            1                          if it takes more than                     5                     flips to get a tails on day                     i                                      0                          otherwise                        Let   A  =      ∑          i      =      1              365            A    i  By definition, A is binomially distributed with   n  =  365 trials and   p  =  P  r  (      A    i    =  1  )What is   P  r  (      A    i    =  1  )? To see this, notice that one of exactly two situations will occur: The first tail is flipped within the first five attempts, or the first five attempts are all heads.Step 2We see then that it doesn&#039;t actually matter to this problem when the first tail is, just what the outcomes of the first five flips are. We see that   P  r  (      A    i    =  1  )  =  P  r  (  five out of five flips all come up heads  )  =  (      1    2        )    5    =      1    32  The problem asks us to calculate   P  r  (  A  ≥  4  )Remembering that   P  r  (  X  =  k  )  =                    (                    n      k                      )                  p    k    (  1  −  p      )          n      −      k       for a binomially distributed random variable, and applying the rule of complementary events, we have:  P  r  (  A  ≥  4  )  =  1  −  P  r  (  A  =  0  )  −  P  r  (  A  =  1  )  −  P  r  (  A  =  2  )  −  P  r  (  A  =  3  )    =  1  −                    (                    365      0                      )                        31              365                    32              365              −                    (                    365      1                      )                        31              364                    32              365              −                    (                    365      2                      )                        31              363                    32              365              −                    (                    365      3                      )                        31              362                    32              365

Probability that at least 4 out of 365 times I try, I need more than 5 flips before getting a tails

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