Question

in a certain college, 33% of the physics majors belong to ethnic minorities. if 10 students are selected at random from the physics majors, what is the probability that no more than 6 belong to an ethnic minority? round to three decimal places.

a. 0.913
b. 0.981
c. 0.985
d. 0.055

Explanation:

Step1: Identify the binomial - probability formula

The binomial - probability formula is $P(X = k)=C(n,k)\times p^{k}\times(1 - p)^{n - k}$, where $n$ is the number of trials, $k$ is the number of successes, $p$ is the probability of success on a single trial, and $C(n,k)=\frac{n!}{k!(n - k)!}$. Here, $n = 10$, $p=0.33$, and $1 - p = 0.67$. We want to find $P(X\leq6)=1-(P(X = 7)+P(X = 8)+P(X = 9)+P(X = 10))$.

Step2: Calculate $P(X = 7)$

$C(10,7)=\frac{10!}{7!(10 - 7)!}=\frac{10!}{7!3!}=\frac{10\times9\times8}{3\times2\times1}=120$.
$P(X = 7)=C(10,7)\times(0.33)^{7}\times(0.67)^{3}=120\times(0.33)^{7}\times(0.67)^{3}\approx120\times0.000426\times0.300763\approx0.0154$.

Step3: Calculate $P(X = 8)$

$C(10,8)=\frac{10!}{8!(10 - 8)!}=\frac{10!}{8!2!}=\frac{10\times9}{2\times1}=45$.
$P(X = 8)=C(10,8)\times(0.33)^{8}\times(0.67)^{2}=45\times(0.33)^{8}\times(0.67)^{2}\approx45\times0.000141\times0.4489\approx0.0028$.

Step4: Calculate $P(X = 9)$

$C(10,9)=\frac{10!}{9!(10 - 9)!}=10$.
$P(X = 9)=C(10,9)\times(0.33)^{9}\times(0.67)^{1}=10\times(0.33)^{9}\times0.67\approx10\times0.000047\times0.67\approx0.0003$.

Step5: Calculate $P(X = 10)$

$C(10,10)=\frac{10!}{10!(10 - 10)!}=1$.
$P(X = 10)=C(10,10)\times(0.33)^{10}\times(0.67)^{0}=(0.33)^{10}\approx0.000013$.

Step6: Calculate $P(X\leq6)$

$P(X = 7)+P(X = 8)+P(X = 9)+P(X = 10)\approx0.0154 + 0.0028+0.0003+0.000013\approx0.0185$.
$P(X\leq6)=1 - 0.0185 = 0.9815\approx0.981$.

Answer:

B. 0.981