Aryanna Blake

Aryanna Blake

Answered

2022-10-21

Probability of two successes, possible geometric distribution problem
Let Y be the number of trials required to observe r successes. List the outcome space for the variable Y and show that the probabilities for Y are giver by the formula P ( Y = k ) = ( k 1 ) ( p 2 ) ( 1 p ) k 2 .
Okay I know the outcome space is Y := 2 , 3 , 4 , . . . , the first two trials could be successes or you could never get two successes. The problem looks geometric but I dont undertand the extra ( k 1 ) p in front.

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Recalculate according to your conditions!

Answer & Explanation

Steinherrjm

Steinherrjm

Expert

2022-10-22Added 12 answers

Step 1
Assuming independence of trials and that each trial has probability p of success and also that r = 2, as in the title:
P [ Y = k ] is the probability that the second success occurred on the k'th trial. If the second success occurred on the k'th trial, then there was exactly one success in the first k 1 trials and the k'th trial was a success.
Now, there are k 1 ways to have exactly one success in the first k 1 trials (for example, one way is that the second trial was a success and the other k 2 trials were failures), each of which having probability ( 1 p ) k 2 p. So the probability that exactly one success occurred in the first k 1 trials is
P(exactly one success in the first (1) P ( exactly one success  in the first  k 1  trials ) = ( k 1 ) ( 1 p ) k 2 p .
Step 2
The probability of a success on the k'th trial is (2) P (  success on the k 'th trial ) = p
The probability that the second success occurred on the k'th trial is the product of (1) and (2):
P [ Y = k ] = ( ( k 1 ) ( 1 p ) k 2 p ) p = ( k 1 ) p 2 ( 1 p ) k 2 .

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