Nicholas J. answered • 08/01/15

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The binomial distribution counts the number of successes in a set of n identical trials. The probability of success, or a '1' on each individual trial is given by p. Therefore,

p = 0.15 corresponds to counting a defective item as a 'success'

p = 0.85 corresponds to counting a non-defective item as a 'success'

Either choice is fine, as long as you are consistent with it throughout the rest of your analysis

For example, if p = 0.15, i.e. if we are counting defectives as successes then

P(Exactly 4 defective) = (5 choose 4)*p^4*(1-p)^1 = 5*0.15^4*0.85

Alternatively, if p = 0.85, i.e. if we are counting non-defectives as successes then

P(Exactly 4 defective) = (5 choose 1)*p^1*(1-p)^4 = 5*0.85*0.15^4

P(Exactly 4 defective) = (5 choose 1)*p^1*(1-p)^4 = 5*0.85*0.15^4

Nicholas J.

The key here is that p represents the probability of success of failure on Each individual trial.

There are n=5 trials, each with probability p = 0.15 of being defective.

I think maybe you are confusing p with the average number of defective items that would be produced in a sample of 5 items, this is also sometimes referred to as the 'expectation' of the binomial distribution.

The formula for the expectation of a binomial distribution is n*p

Or in this case, the expected number of defective items is 5*0.15 = 0.75

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08/01/15

Nicholas J.

Another interesting thing to point out is that the expectation n*p is not necessarily less than or equal to 1.

p is the probability of a 'success' on each trial,

Because p is a probability, we know that p must be less than or equal to 1.

For example, if we were looking at a sample of size 50, instead a sample of size 5, then

n*p = 50*0.15 = 7.5

but we can't set p = 7.5, because then p wouldn't be a valid probability

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08/01/15

Koshila D.

08/01/15