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# Statistics Question (Probability)?

A pharmaceutical company receives large shipments of ibuprofen tablets and uses this acceptance sampling plan: randomly select and test 23 tablets, then accept the whole batch if there is at most one that doesn’t meet the required specifications. If a particular shipment of thousands of ibuprofen tablets actually has a 12% rate of defects, what is the probability that this whole shipment will be accepted?

(Report answer as a decimal value accurate to four decimal places.)

P(accept shipment) =______?

Please help and explain! thanks

### 2 Answers

- 8 years agoFavorite Answer
First, let Y denote the number of defective tablets in the 23

So Y ~ Binomial(n=23, p=0.12)

Then,

P(accept shipment) = P(Y≤1)=P(Y=0)+P(Y=1)

We have,

P(Y=0)= (23 choose 0)*(0.12)^0*(1-0.12)^23=0.05286

P(Y=1)= (23 choose 1)*(0.12)^1*(1-0.12)^22=0.16578

So,

P(Y≤1)=0.21864

This should be the answer.

What we have here is a binomial distribution with 23 Bernoulli trials, each having a probability of 0.12 of occuring.

You want at most one defective tablet, so both 0 and 1 defective tablets is o.k.

Then you apply the probability formula of a binomial distribution to both of these outcomes.

Generally,

P(Y=y)=(n choose y)*p^y*(1-p)^(n-y)

In this case, n=23, p=0.12

Look at it this way, (n choose y) ensures that all the permutations are taken into count, of the 23, y are defective and 23-y are not defective.

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- Anonymous8 years ago
Ans:

Let x be the random variable denotes the no. of defective tablets.

Need to find P(x<=1)=?

It follows binomial distribution with n=12, p=0.12 and q=1-0.12=0.88

P(x<=1)=P(x=0)+P(x=1)

= 12c0*(0.12)^0*(0.88)^12 + 12c1*(0.12)^1*(0.88)^11

=0.5685

P(accept shipment) =P(x<=1)=0.5685(Ans.)

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