Question

Each sample of water has a 30% chance of containing a particular organic pollutant. Assume that the samples are independent with regard to the presence of the pollutant. Find the probability that, in the next 10 samples, two or less contain the pollutant. (Hint: using the tables in Appendix A will be faster than doing it by hand!)
a) 0.678
b) 0.028
c) 0.041
d) 0.383

Answer 1

Answer:

d) 0.383

Step-by-step explanation:

For each sample, there are only two possible outcomes. Either they contain the pollutant, or they do not. The probability of a sample containing the pollutant is independent of any other sample. Thus, the binomial probability distribution is used to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

30% chance of containing a particular organic pollutant.

This means that $p = 0.3$

Next 10 samples

This means that $n = 10$

Probability that two or less contain the pollutant.

This is:

$P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2)$. So

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{10,0}.(0.3)^{0}.(0.7)^{10} = 0.028$

$P(X = 1) = C_{10,1}.(0.3)^{1}.(0.7)^{9} = 0.121$

$P(X = 2) = C_{10,2}.(0.3)^{2}.(0.7)^{8} = 0.233$

$P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2) = 0.028 + 0.121 + 0.233 = 0.382$

A little rounding difference, but the correct answer is given by option d.