Answer to Question #185948 in Statistics and Probability for Sajid

Question

Q(1) [10 Marks] [CLO2,C3] (a) 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 PMF of X. Find P(X > 15). (b) Packets at a certain node on the internet arrive with a rate of 100 packets per minute. Find the probability that no packets arrive in 6 seconds. Find the probability that 2 or more packets arrive in the first 6 seconds.

Answer 1

Solution (a):

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 $\frac{1}{4}$ . Hence, you perform 10 independent $\operatorname{Bernoulli}\left(\frac{1}{4}\right)$ trials and Y is the number of successes. Thus, we conclude $Y \sim Binomial \left(10, \frac{1}{4}\right)$ , so

$P_{Y}(y)=\left\{\begin{array}{ll} \left(\begin{array}{c} 10 \\ y \end{array}\right)\left(\frac{1}{4}\right)^{y}\left(\frac{3}{4}\right)^{10-y} & \text { for } y=0,1,2,3, \ldots, 10 \\ 0 & \text { otherwise } \end{array}\right.$

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

$\begin{aligned} P_{X}(10) &=P(X=10)=P(Y+10=10) \\ &=P(Y=0)=\left(\begin{array}{c} 10 \\ 0 \end{array}\right)\left(\frac{1}{4}\right)^{0}\left(\frac{3}{4}\right)^{10-0}=\left(\frac{3}{4}\right)^{10} \\ P_{X}(11) &=P(X=11)=P(Y+10=11) \\ &=P(Y=1)=\left(\begin{array}{c} 10 \\ 1 \end{array}\right)\left(\frac{1}{4}\right)^{1}\left(\frac{3}{4}\right)^{10-1}=10\left(\frac{1}{4}\right)\left(\frac{3}{4}\right)^{9} . \end{aligned}$

In general for $k \in R_{X}=\{10,11,12, \ldots, 20\}$ ,

$\begin{aligned} P_{X}(k) &=P(X=k)=P(Y+10=k) \\ &=P(Y=k-10)=\left(\begin{array}{l} 10 \\ k-10 \end{array}\right)\left(\frac{1}{4}\right)^{k-10}\left(\frac{3}{4}\right)^{20-k} \end{aligned}$

Thus,

$P_{X}(k)=\left\{\begin{array}{ll} \left(\begin{array}{l} 10 \\ k-10 \end{array}\right)\left(\frac{1}{4}\right)^{k-10}\left(\frac{3}{4}\right)^{20-k} & \text { for } k=10,11,12, \ldots, 20 \\ 0 & \text { otherwise } \end{array}\right.$

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

$\begin{array}{c} P_{Y}(y)=\left\{\begin{array}{ll} \left(\begin{array}{c} 10 \\ y \end{array}\right)\left(\frac{1}{4}\right)^{y}\left(\frac{3}{4}\right)^{10-y} & \text { for } y=6,7,8,9,10 \\ 0 & \text { otherwise } \end{array}\right. \\ P_{X}(k)=\left\{\begin{array}{ll} \left(\begin{array}{c} 10 \\ k-10 \end{array}\right)\left(\frac{1}{4}\right)^{k-10}\left(\frac{3}{4}\right)^{20-k} & \text { for } k=16,17, \ldots, 20 \\ 0 & \text { otherwise } \end{array}\right. \\ P(X>15)=P_{X}(16)+P_{X}(17)+P_{X}(18)+P_{X}(19)+P_{X}(20) \\ =\left(\begin{array}{c} 10 \\ 6 \end{array}\right)\left(\frac{1}{4}\right)^{6}\left(\frac{3}{4}\right)^{4}+\left(\begin{array}{c} 10 \\ 7 \end{array}\right)\left(\frac{1}{4}\right)^{7}\left(\frac{3}{4}\right)^{3}+\left(\begin{array}{c} 10 \\ 8 \end{array}\right)\left(\frac{1}{4}\right)^{8}\left(\frac{3}{4}\right)^{2} \\ +\left(\begin{array}{c} 10 \\ 9 \end{array}\right)\left(\frac{1}{4}\right)^{9}\left(\frac{3}{4}\right)^{1}+\left(\begin{array}{c} 10 \\ 10 \end{array}\right)\left(\frac{1}{4}\right)^{10}\left(\frac{3}{4}\right)^{0} \end{array}$

$=0.019727$

(b):

Given, $\lambda=100$ per minute$=\dfrac{100}{60}\ or\ \dfrac{5}{3}$ per second

For 6 second, $\lambda=\dfrac{5}{3}\times 6=10$

$P(X=0)=10^0\dfrac{e^{-10}}{0!}=0.00004539$

Next, $P(X\ge2)=1-P(X<2)=1-[P(X=0+P(X=1)]$

$=1-(0.00004539+10^1\dfrac{e^{-10}}{1!})=0.9995006$