Probability distribution binomial. An assembly system is composed of n independent and identical parts. During any given “run” of the system, all parts have a probability of p of working. Suppose the random variable Y represents the number of working parts in any given “run”. You are told that P(Y=3) = 2P(Y=2) and V[Y] = p. What is the probability distribution of Y?

metal1fc

metal1fc

Answered question

2022-09-14

Probability distribution binomial
An assembly system is composed of n independent and identical parts. During any given “run” of the system, all parts have a probability of p of working. Suppose the random variable Y represents the number of working parts in any given “run”. You are told that P ( Y = 3 ) = 2 P ( Y = 2 ) and V [ Y ] = p. What is the probability distribution of Y?
I know thus is a binomial wherein the mean is np and the variance is np(1-p) how would you solve this?

Answer & Explanation

faliryr

faliryr

Beginner2022-09-15Added 15 answers

Step 1
There is the trivial solution p = 0, in which case n is cannot be determined. Assume now that p 0.
We have Pr ( Y = 3 ) = ( n 3 ) p 3 ( 1 p ) n 3 and Pr ( Y = 2 ) = ( n 2 ) p 2 ( 1 p ) n 2
We were told that
( n 3 ) p 3 ( 1 p ) n 3 = 2 ( n 2 ) p 2 ( 1 p ) n 2 .
Thus
n ( n 1 ) ( n 2 ) 6 p 3 ( 1 p ) n 3 = 2 n ( n 1 ) 2 p 2 ( 1 p ) n 2 .
Step 2
A bit of algebra simplifies this to
(1) ( n 2 ) p = 6 ( 1 p ) .
The information about the variance tells us that n ( 1 p ) = 1
Solve for p, by replacing n in (1) by 1 1 p . We get p = 3 4 and then n = 4.

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