Question

mrs. gomes found that 40% of students at her high school take chemistry. she randomly surveys 12 students. what is the probability that exactly 4 students have taken chemistry? round the answer to the nearest thousandth.

$p(k\text{ successes}) = _nc_kp^k(1 - p)^{n - k}$

$_nc_k=\frac{n!}{(n - k)!k!}$

0.005
0.008
0.213
0.227

Explanation:

Step1: Identify values of n, k, p

n = 12 (number of students surveyed), k = 4 (number of students taking chemistry), p = 0.4 (probability of a student taking chemistry)

Step2: Calculate combination \(_{n}C_{k}\)

\[_{12}C_{4}=\frac{12!}{(12 - 4)!4!}=\frac{12!}{8!4!}=\frac{12\times11\times10\times9}{4\times3\times2\times1}=495\]

Step3: Calculate \(p^{k}\) and \((1 - p)^{n - k}\)

\(p^{k}=(0.4)^{4}=0.0256\), \((1 - p)^{n - k}=(1 - 0.4)^{12 - 4}=(0.6)^{8}=0.01679616\)

Step4: Calculate probability \(P(k)\)

\[P(4)=_{12}C_{4}\times p^{4}\times(1 - p)^{8}=495\times0.0256\times0.01679616\approx0.213\]

Answer:

0.213