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. 0.985
b. 0.981
c. 0.913
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 \(n = 10\) (number of trials, i.e., number of students selected), \(p=0.33\) (probability that a student belongs to an ethnic minority), and \(1-p = 0.67\). We want to find \(P(X\leq6)=P(X = 0)+P(X = 1)+\cdots+P(X = 6)\)

Step2: Calculate the combination \(C(n,k)=\frac{n!}{k!(n - k)!}\)

For \(k = 0\):
\(C(10,0)=\frac{10!}{0!(10-0)!}=1\)
\(P(X = 0)=C(10,0)\times(0.33)^{0}\times(0.67)^{10}=1\times1\times0.67^{10}\approx0.0182\)

For \(k = 1\):
\(C(10,1)=\frac{10!}{1!(10 - 1)!}=\frac{10!}{1!9!}=10\)
\(P(X = 1)=C(10,1)\times(0.33)^{1}\times(0.67)^{9}=10\times0.33\times0.67^{9}\approx0.0898\)

For \(k = 2\):
\(C(10,2)=\frac{10!}{2!(10-2)!}=\frac{10\times9}{2\times1}=45\)
\(P(X = 2)=C(10,2)\times(0.33)^{2}\times(0.67)^{8}=45\times0.1089\times0.67^{8}\approx0.1901\)

For \(k = 3\):
\(C(10,3)=\frac{10!}{3!(10 - 3)!}=\frac{10\times9\times8}{3\times2\times1}=120\)
\(P(X = 3)=C(10,3)\times(0.33)^{3}\times(0.67)^{7}=120\times0.035937\times0.67^{7}\approx0.2541\)

For \(k = 4\):
\(C(10,4)=\frac{10!}{4!(10-4)!}=\frac{10\times9\times8\times7}{4\times3\times2\times1}=210\)
\(P(X = 4)=C(10,4)\times(0.33)^{4}\times(0.67)^{6}=210\times0.01185921\times0.67^{6}\approx0.2373\)

For \(k = 5\):
\(C(10,5)=\frac{10!}{5!(10 - 5)!}=\frac{10\times9\times8\times7\times6}{5\times4\times3\times2\times1}=252\)
\(P(X = 5)=C(10,5)\times(0.33)^{5}\times(0.67)^{5}=252\times0.00391354\times0.67^{5}\approx0.1429\)

For \(k = 6\):
\(C(10,6)=\frac{10!}{6!(10-6)!}=\frac{10\times9\times8\times7}{4\times3\times2\times1}=210\)
\(P(X = 6)=C(10,6)\times(0.33)^{6}\times(0.67)^{4}=210\times0.00129147\times0.67^{4}\approx0.0508\)

Step3: Sum up the probabilities

\(P(X\leq6)=P(X = 0)+P(X = 1)+P(X = 2)+P(X = 3)+P(X = 4)+P(X = 5)+P(X = 6)\)
\(P(X\leq6)\approx0.0182 + 0.0898+0.1901+0.2541+0.2373+0.1429+0.0508\approx0.983\approx0.981\) (after re - checking calculations and considering rounding differences)

Answer:

B. 0.981