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(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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    @Gigili There were actually three. During first couple of minutes after the question was asked.2012-06-07

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The first question was completely answered by sam in the comments.

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 $30$%.