Probability of multiple variables, geometric distribution?You are on a basketball team, and at the end of every practice, you shoot half-court shots until you make one. Once you make a shot, you go home. Each half-court shot, independent of all other shots, has a 0.1 probability of going in. Your team has 100 practices per season. Estimate the probability that you shoot more than 1111 half-court shots after practices this season.So I set X equal to the number of shots until finished per practice, and since X ∼ G e o ( 0.1 ), I know that the P(X) is ( 1 − p ) x − 1 p but I don't know what to set x as.I also set Y equal to the number of shots until finished per season so that Y = 100 X. I'm not sure how to go about finding the probability though? I understand that P ( Y ) = P ( X ) 100 , but how can I find P(X)?

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Probability of multiple variables, geometric distribution?You are on a basketball team, and at the end of every practice, you shoot half-court shots until you make one. Once you make a shot, you go home. Each half-court shot, independent of all other shots, has a 0.1 probability of going in. Your team has 100 practices per season. Estimate the probability that you shoot more than 1111 half-court shots after practices this season.So I set X equal to the number of shots until finished per practice, and since   X  ∼      G    e    o    (  0.1  ), I know that the P(X) is   (  1  −  p      )          x      −      1        p but I don&#039;t know what to set x as.I also set Y equal to the number of shots until finished per season so that   Y  =  100  X. I&#039;m not sure how to go about finding the probability though? I understand that   P  (  Y  )  =  P  (  X      )          100      , but how can I find P(X)?

Steven Bates · Accepted Answer

Step 1Outline Let       X    1    ,      X    2    ,  …  ,      X    n  , where   n  =  100, be the number of shots taken on practice   1  ,  2  ,  …  ,  n.Let   Y  =      X    1    +      X    2    +  ⋯  +      X    n  . We want the probability that   Y  &amp;gt;  1111.Each       X    i   has geometric distribution. You know the mean of       X    i  . Call it   μ. Find or look up what the variance is. Call it       σ    2  .By the Central Limit Theorem, Y has roughly normal distribution, with mean   n  μ and variance   n      σ    2  . Now you can calculate.Step 2Remark: The post refers to 100X. This is an entirely different sort of random variable than our random variable Y, and its distribution is not relevant to our problem.You asked about the distribution of Y. This random variable measures the number of shots until the 100-th success. It has negative binomial distribution. It is not hard to write down an explicit formula for the probability that   Y  =  y.Then we could solve our problem by finding the sum of   Pr  (  Y  =  y  ), from   y  =  100 to   y  =  1111, and subtracting the result from 1. This involves an unpleasant amount of computation, though the right kind of software could handle it. That&#039;s the reason that we find the normal approximation path more appealing.

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