Probability distribution for a geometric distribution don't add up to 1Say I'm rolling 2 dies, numbered 1 to 10. A successful outcome is considered rolling a multiple of 4. Therefore, probability of success = 0.25 and prob of failure = 0.75. This is an example of a geometric distribution. I can roll a maximum of 6 times. I made the prob distribution chart to represent all the different possibilities, but my values don't add up to 1. I don't understand what I'm doing wrong. Roll Number Probability X = 1 ( 0.75 ) ˆ 0 ( 0.25 ) 0.25 X = 2 ( 0.75 ) ˆ 1 ( 0.25 ) 0.1875 X = 3 ( 0.75 ) ˆ 2 ( 0.25 ) 0.140625 X = 4 ( 0.75 ) ˆ 3 ( 0.25 ) 0.10546875 X = 5 ( 0.75 ) ˆ 4 ( 0.25 ) 0.079101562 X = 6 ( 0.75 ) ˆ 5 ( 0.25 ) 0.059326171

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Probability distribution for a geometric distribution don&#039;t add up to 1Say I&#039;m rolling 2 dies, numbered 1 to 10. A successful outcome is considered rolling a multiple of 4. Therefore, probability of success   =  0.25 and prob of failure   =  0.75. This is an example of a geometric distribution. I can roll a maximum of 6 times. I made the prob distribution chart to represent all the different possibilities, but my values don&#039;t add up to 1. I don&#039;t understand what I&#039;m doing wrong.                    Roll                     Number                          Probability                                        X                            =        1                                  (        0.75        )                  ˆ                0        (        0.25        )                          0.25                            X                            =        2                                  (        0.75        )                  ˆ                1        (        0.25        )                          0.1875                            X                            =        3                                  (        0.75        )                  ˆ                2        (        0.25        )                          0.140625                            X                            =        4                                  (        0.75        )                  ˆ                3        (        0.25        )                          0.10546875                            X                            =        5                                  (        0.75        )                  ˆ                4        (        0.25        )                          0.079101562                            X                            =        6                                  (        0.75        )                  ˆ                5        (        0.25        )                          0.059326171

cenjene9gw · Accepted Answer

Step 1Since the random variable X has not been precisely defined, we will supply a (speculative) definition.The game consists of rolling the two dice repeatedly. We stop when we roll a sum divisible by 4 or when the total number of rolls reaches 6. Let X be the number of rolls. It is indeed true that the probability of getting a sum divisible by 4 on a roll of the two dice (probability of a success) is 0.25.Step 2For   k  =  1 to 5, the probability that   X  =  k is given by the familiar geometric distribution. Thus   Pr  (  X  =  k  )  =  (  0.75      )          k      −      1        (  0.25  ).But   k  =  6 is special. We have   k  =  6 precisely if there have been 5 failures in a row. Thus   Pr  (  X  =  6  )  =  (  0.75      )    5  .

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