Question

In Quebec, 90 percent of the population subscribes to the Roman Catholic religion. In a random sample of eight Quebecois, find the probability that the sample contains at least five Roman Catholics.

Answer 1

Answer:

Probability that the sample contains at least five Roman Catholics = 0.995 .

Step-by-step explanation:

We are given that In Quebec, 90 percent of the population subscribes to the Roman Catholic religion.

The Binomial distribution probability is given by;

 P(X = r) = $\binom{n}{r}p^{r}(1-p)^{n-r}$ for x = 0,1,2,3,.......

Here, n = number of trials which is 8 in our case

         r = no. of success which is at least 5 in our case

         p = probability of success which is probability of Roman Catholic of

                 0.90 in our case

So, P(X >= 5) = P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8)

= $\binom{8}{5}0.9^{5}(1-0.9)^{8-5} + \binom{8}{6}0.9^{6}(1-0.9)^{8-6} + \binom{8}{7}0.9^{7}(1-0.9)^{8-7} + \binom{8}{8}0.9^{8}(1-0.9)^{8-8}$

= 56 * $0.9^{5} * (0.1)^{3}$ + 28 * $0.9^{6} * (0.1)^{2}$ + 8 * $0.9^{7} * (0.1)^{1}$ + 1 * $0.9^{8}$

= 0.995

Therefore, probability that the sample contains at least five Roman Catholics is 0.995.