Answer to Question #323842 in Statistics and Probability for Bless

Question

 You take an exam that contains 20 multiple-choice questions. Each question has 4 possible options. You know the answer to 10 questions, but you have no idea about the other 10 questions so you choose answers randomly. Your score X on the exam is the total number of correct answers. Find the p.m.f. of X. What is P(X > 15)?

Answer 1

Let's define the random variable Y as the number of your correct answers to the 10 questions you answer randomly. Then your total score will be X=Y+10. First, let's find the PMF of Y. For each question, your success probability is 1/4. Hence, you perform 10 independent Bernoulli 1/4 trials and Y is the number of successes. Thus, we conclude

$P_y=(y=(^{10}_{y})(1/4)^y(3/4)^{10-y}$

for y=0,1,2,3...10

$P_y=0$ otherwise

Now we need to find the PMF of X=Y+10. First note that $R_x=\{10,11,12...20\}$

We can write

$P_x(10)=P(x=10)=P(y+10=10)=P(y=0)=(^{10}_{0})(1/4)^0(3/4)^{10-0}=(3/4)^{10}\\P(x=11)=P(y+10=11)=P(y=1)=(^{10}{1})(1/4)^1(3/4)^{10-1}=10(1/4)(3/4)^9$

In general for

$k \isin R_x=\{10,11,12...20\}$

$P_x(k)=P(x=k)=P(y+10=k)=P(y=k-10)=(^{10}_{k-10})(1/4)^{k-10}(3/4)^{20-k}$

For k={10,11,12...20}

$P_x(k)=0$

Otherwise

In order to calculate P(X>15), we know we should consider y=6,7,8,9,10

$P_y(y)=(^{10}_y)(1/4)^y(3/4)^{10-y}$

For y=6,7,8,9,10

P_y(y)=0 otherwise

$P_x(k)=(^{10}_{k-10})(1/4)^{k-10}(3/4)^{20-k}$

For k=16,17,18,19,20

$P_x(k)=0$ otherwise

$P(X>15)=P_x(16)+P_x(17)+P_x(18)+P_x(19)+P_x(20)=(^{10}_6)(1/4)^6(3/4)^4+(^{10}_7)(1/4)^7(3/4)^3+(^{10}_8)(1/4)^8(3/4)^2+(^{10}_9)(1/4)^9(3/4)^1+(^{10}_{10})(1/4)^{10}(3/4)^0=0.019727$