# Binomial probability distribution

(1) Three items are selected at random from a manufacturing process & classified defective or non-defective. A defective item is designated a success. Assume $25\%$ of the production is defective. Let $X$ be the number of successes, let $N$: non-defective, $D$: defective.

When I use the formula $P(X=2)=b(2;3,0.25)=9/64$. But when I use $P(NDD)=(3/4)(1/4)(1/4)=3/64$. What is wrong here?

(2) In a certain rural community "$30\%$ of wells are impure" is merely a conjecture put forth by the area water board.Suppose $10$ wells are randomly selected & $6$ are found to contain the impurity.

By using the formula $P(X=6)=0.0367$, please explain which result you compare with the result $0.0367$ to conclude that the conjecture is unlikely.

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THANK YOU FOR SAYING "PLEASE" – The Chaz 2.0 Jun 7 '12 at 5:21
YOUR QUESTION IS VERY HARD TO UNDERSTAND. PLEASE CLARIFY AND PROVIDE CONTEXT TO RECEIVE A PROPER ANSWER. – Zolani13 Jun 7 '12 at 5:22
I think it does not look nice that OP gets so many unhelpful comments. I am sure there is a better place to exercise somebody's humor. – Artem Jun 7 '12 at 5:26
@Artem You're right; my apologies. – Zolani13 Jun 7 '12 at 5:27
For (1), you are using the binomial distribution to find the chance of two non defective items with one defective item. On the other hand, $P(NND)=P(N)P(D)P(D)$ calculates the probability that the first two items are not defective and the third is defective ("order matters"). So to get the same answer as the binomial distribution, you need to add the possibility of DND and DDN. – Alex R. Jun 7 '12 at 5:30

The second question isn’t about calculating anything, but rather about interpreting a statistic. If the water board’s conjecture is correct, then the probability that a randomly chosen well is impure is $0.3$. When we select $10$ wells at random and test them for impurity, letting $X$ be the number of impure wells, then $X$ follows a binomial distribution with $n=10$ and $p=0.3$. The probability of finding $6$ impure wells is then $P(X=6)=b(6;10,0,3)$, which is actually $0.0368$ when correctly rounded to $4$ decimal places. In other words if the water board’s conjecture is correct, this particular sample of $10$ wells is extremely improbable: only about $3.68$% of all random samples of $10$ wells from the area should have $6$ impure wells. Either we just happened to pick a very unusual sample at random, or the water board’s conjecture is too optimistic, and the actual percentage of impure wells is more than $30$%.