Question

Suppose that 16 percent of the students in a certain high school are freshmen, 14 percent are sophomores, 38 percent are juniors, and 32 percent are seniors. If 15 students are selected at random from the school, what is the

The probability that at least eight will be either freshmen or sophomores?

Short answer

0.0501

Step-by-step solution

Given information

It is given that

\(\begin{array}{l}{p_1} = 0.16\\{p_2} = 0.14\\{p_3} = 0.38\\{p_4} = 0.32\end{array}\)

Which denotes the percent of students in freshman, sophomores, juniors, and seniors, respectively.

Define the pdf

This is a case of a multinomial distribution.

The pdf of a standard multinomial distribution is:

\(\)\(f\left( {x|n,p} \right) = \left\{ \begin{array}{l}\frac{{n!}}{{{x_1}! \ldots {x_n}!}}{p_1}^{{x_1}} \ldots {p_k}^{{x_k}},if\,{x_1} + \ldots + {x_k} = n\\0,otherwise\end{array} \right.\)

In this case, n=15

\(\frac{{15!}}{{{x_1}!{x_2}!{x_3}!{x_4}!}}{\left( {0.16} \right)^{{x_1}}}{\left( {0.14} \right)^{{x_2}}}{\left( {0.38} \right)^{{x_3}}}{\left( {0.32} \right)^{{x_4}}}\)

Calculate the required probability

We have to calculate the probability that at least eight will be either freshmen or sophomores.

1) \(P\left( {{X_1} > 8} \right)\)

Therefore, we combine sophomore, junior and senior probabilities in one.

\(\begin{array}{l} = \frac{{15!}}{{{x_1}!{x_2}!}}{\left( {0.16} \right)^{{x_1}}}{\left( {0.84} \right)^{{x_2}}}\\ = \frac{{15!}}{{8!7!}}{\left( {0.16} \right)^8}{\left( {0.84} \right)^7} + \frac{{15!}}{{9!6!}}{\left( {0.16} \right)^9}{\left( {0.84} \right)^6} + \frac{{15!}}{{10!5!}}{\left( {0.16} \right)^{10}}{\left( {0.84} \right)^5} + \ldots + \frac{{15!}}{{15!}}{\left( {0.16} \right)^{15}}\\\left( 1 \right)\end{array}\)

2) Similarly, \(P\left( {{X_2} > 8} \right)\)

Therefore, we combine freshman, junior and senior probabilities in one.

\(\begin{array}{l} = \frac{{15!}}{{{x_2}!{x_1}!}}{\left( {0.14} \right)^{{x_2}}}{\left( {0.86} \right)^{{x_1}}}\\ = \frac{{15!}}{{8!7!}}{\left( {0.14} \right)^8}{\left( {0.86} \right)^7} + \frac{{15!}}{{9!6!}}{\left( {0.14} \right)^9}{\left( {0.86} \right)^6} + \frac{{15!}}{{10!5!}}{\left( {0.14} \right)^{10}}{\left( {0.86} \right)^5} + \ldots + \frac{{15!}}{{15!}}{\left( {0.14} \right)^{15}}\\\left( 2 \right)\end{array}\)

Note, here \({x_1}\,\)is the combined probability of all except sophomores.

Adding (1) and (2)

0.0501

The final answer is 0.0501