mrs. gomes found that 40% of students at her high school take chemistry. she randomly surveys 12 students…

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.\n\n$p(k\text{ successes}) = _nc_kp^k(1 - p)^{n - k}$\n\n$_nc_k=\frac{n!}{(n - k)!k!}$\n\n0.008\n0.438\n0.562\n0.665

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.\n\n$p(k\text{ successes}) = _nc_kp^k(1 - p)^{n - k}$\n\n$_nc_k=\frac{n!}{(n - k)!k!}$\n\n0.008\n0.438\n0.562\n0.665

Answer

Answer:

B. 0.438

Explanation:

Step1: Identify values

$n = 12$, $p=0.4$, $1 - p = 0.6$

Step2: Calculate probabilities for $k = 0,1,2,3,4$

$P(X = k)=\binom{n}{k}p^{k}(1 - p)^{n - k}$ For $k = 0$: $\binom{12}{0}(0.4)^{0}(0.6)^{12}=\frac{12!}{0!(12 - 0)!}\times1\times(0.6)^{12}= 0.002176782336$ For $k = 1$: $\binom{12}{1}(0.4)^{1}(0.6)^{11}=\frac{12!}{1!(12 - 1)!}\times0.4\times(0.6)^{11}=0.01264531424$ For $k = 2$: $\binom{12}{2}(0.4)^{2}(0.6)^{10}=\frac{12!}{2!(12 - 2)!}\times0.16\times(0.6)^{10}=0.0420643808$ For $k = 3$: $\binom{12}{3}(0.4)^{3}(0.6)^{9}=\frac{12!}{3!(12 - 3)!}\times0.064\times(0.6)^{9}=0.10090320384$ For $k = 4$: $\binom{12}{4}(0.4)^{4}(0.6)^{8}=\frac{12!}{4!(12 - 4)!}\times0.0256\times(0.6)^{8}=0.18018048$

Step3: Sum probabilities

$P(X\leq4)=P(X = 0)+P(X = 1)+P(X = 2)+P(X = 3)+P(X = 4)$ $P(X\leq4)=0.002176782336+ 0.01264531424+0.0420643808+0.10090320384+0.18018048\approx0.438$