# It is conjectured that an impurity exists in 30% of all drinking wells in a certain rural...

## Question:

It is conjectured that an impurity exists in 30% of all drinking wells in a certain rural community. In order to gain some insight into the true extent of the problem, it is determined that some testing is necessary. It is too expensive to test all the wells in the area, so 10 are randomly selected for testing.

What is the probability that exactly 3 wells have the impurity, assuming that the conjecture is correct? What is the probability that more than 3 wells are impure?

Suppose 6 wells are found to contain the impurity. What does this imply about the conjecture?

## Unusual Event:

When the probability of an event is less than or equal to {eq}0.05 {/eq} i.e. it occurs less of equal to {eq}5 {/eq} times in {eq}100 {/eq} trials, the event is said to be rare or unusual event. These kinds of events occur rarely and by chance.

The proportion of the drinking wells that contain an impurity, {eq}p = 0.30 {/eq}

The sample size, {eq}n = 10 {/eq}

As per the binomial distribution:

{eq}P(X=r)=^{n}C_{r}\cdot p^{r}\cdot (1-p)^{n-r} {/eq}

The probability that exactly 3 wells have the impurity:

{eq}\begin{align*} P(X=3) & =^{10}C_{3}\times 0.3^{3}\times (1-0.3)^{10-3}\\[1ex] & = 0.2668 \end{align*} {/eq}

The probability that more than 3 wells are impure:

{eq}\begin{align*} P(X > 3) & = 1 - P(X \leq 2)\\[1ex] & = 1 - P(X = 0) - P(X = 1) - P(X = 2)\\[1ex] & = 1 - ^{10}C_{3}\times 0.3^{3}\times (1-0.3)^{10-3} - ^{10}C_{1}\times 0.3^{1}\times (1-0.3)^{10-1} - ^{10}C_{2}\times 0.3^{2}\times (1-0.3)^{10-2}\\[1ex] & = 0.6172 \end{align*} {/eq}

The probability that 6 wells contain the impurity:

{eq}\begin{align*} P(X=6) & =^{10}C_{6}\times 0.3^{6}\times (1-0.3)^{10-6}\\[1ex] & = 0.0368 < 0.0500 \end{align*} {/eq}

As the probability is less than {eq}0.05 {/eq} which represents that it is ususual or rare event. So, the conjecture that there are 6 wells that contain the impurity is not practical.