Answer to Question #320501 in Statistics and Probability for Klum

Question

A student takes a test consisting of 20 true-false questions.

a) what is the probability that the student answers at least seven questions correctly?

b) what is the probability that the student answers at most three questions correctly?

c) what is the probability that the student answers at least four and at most six questions correctly

Answer 1

This is answered by using the binomial probability distribution. If the probability of getting each question right is p, and the total number of questions is n, then the probability of getting exactly r questions right is

a. Probability of 1 correct answer is equal to probability of 1 uncorrect answer and equal p= 0.5, n=20,r=7

$P(X =r)=\frac{n!}{r!(n-r)!}p^r(1-p)^{n-r}$

$P(X\ge7)=1-P(X<7)$

$P(X =0)=\frac{20!}{0!(20)!}0.5^0(0.5)^{20}=9.5\times10^{-7}$

$P(X =1)=\frac{20!}{1!(19)!}0.5^1(0.5)^{19}=1.9\times10^{-5}$

$P(X =2)=\frac{20!}{2!(18)!}0.5^2(0.5)^{18}=0.00018$

$P(X =3)=\frac{20!}{3!(17)!}0.5^3(0.5)^{17}=0.001$

$P(X =4)=\frac{20!}{4!(16)!}0.5^4(0.5)^{16}=0.0046$

$P(X =5)=\frac{20!}{5!(15)!}0.5^5(0.5)^{15}=0.0148$

$P(X =6)=\frac{20!}{6!(14)!}0.5^6(0.5)^{14}=0.037$

$P(X \ge7)=1-0.037-0.0148-0.0046-0.001-0.00018-1.9\times10^{-5}-9.5\times10^{-7}=0.9424$

b. $P(X \le3)=P(X=0)+P(X=1)+P(X=2)+P(X=3)=9.5\times 10^{-7}+1.9\times10^{-5}+0.00018+0.001=0.0012$

c.$P(4 \le X \le6)=P(X=4)+P(X=5)+P(X=6)=0.0046+0.0148+0.037=0.0564$