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

inurbandojoa

inurbandojoa

Answered question

2022-11-08

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

Answer & Explanation

mentest91k99

mentest91k99

Beginner2022-11-09Added 17 answers

Step 1
Let 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 2
We see then that it doesn'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

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