Question

mrs gomes found that 40% of students at her high school take chemistry. she randomly surveys 12 students. what is the probability that at most 4 students have taken chemistry? round the answer to the nearest thousandth.
$p(k\text{ successes}) = _nc_kp^k(1 - p)^{n - k}$
$_nc_k=\frac{n!}{(n - k)!k!}$
0 006
0 438
0 562
0 665

Explanation:

Step1: Identify the binomial - distribution parameters

We have \(n = 12\) (number of trials/surveyed students), \(p=0.4\) (probability of success - student taking chemistry), and we want to find \(P(X\leq4)\). The binomial - probability formula is \(P(X = k)=_{n}C_{k}p^{k}(1 - p)^{n - k}\), where \(_{n}C_{k}=\frac{n!}{(n - k)!k!}\).

Step2: Calculate \(P(X = 0)\)

\(_{12}C_{0}=\frac{12!}{(12-0)!0!}=1\), \(p = 0.4\), \(1 - p=0.6\), \(k = 0\). So \(P(X = 0)=_{12}C_{0}(0.4)^{0}(0.6)^{12}=1\times1\times(0.6)^{12}\approx0.0022\).

Step3: Calculate \(P(X = 1)\)

\(_{12}C_{1}=\frac{12!}{(12 - 1)!1!}=\frac{12!}{11!1!}=12\), \(P(X = 1)=_{12}C_{1}(0.4)^{1}(0.6)^{11}=12\times0.4\times(0.6)^{11}\approx0.0174\).

Step4: Calculate \(P(X = 2)\)

\(_{12}C_{2}=\frac{12!}{(12 - 2)!2!}=\frac{12\times11}{2\times1}=66\), \(P(X = 2)=_{12}C_{2}(0.4)^{2}(0.6)^{10}=66\times0.16\times(0.6)^{10}\approx0.0639\).

Step5: Calculate \(P(X = 3)\)

\(_{12}C_{3}=\frac{12!}{(12 - 3)!3!}=\frac{12\times11\times10}{3\times2\times1}=220\), \(P(X = 3)=_{12}C_{3}(0.4)^{3}(0.6)^{9}=220\times0.064\times(0.6)^{9}\approx0.1419\).

Step6: Calculate \(P(X = 4)\)

\(_{12}C_{4}=\frac{12!}{(12 - 4)!4!}=\frac{12\times11\times10\times9}{4\times3\times2\times1}=495\), \(P(X = 4)=_{12}C_{4}(0.4)^{4}(0.6)^{8}=495\times0.0256\times(0.6)^{8}\approx0.2128\).

Step7: Calculate \(P(X\leq4)\)

\(P(X\leq4)=P(X = 0)+P(X = 1)+P(X = 2)+P(X = 3)+P(X = 4)\)
\(P(X\leq4)\approx0.0022 + 0.0174+0.0639+0.1419+0.2128\approx0.438\).

Answer:

0.438