Virtual Labs

Discrete and Continuous Random Variables

1. I roll a fair die repeatedly until a number less than 2 is observed. If

N

is the total number of times I roll the die, find

P(N = k)

for

k = 1, 2, 3, \ldots

P(N = k) = \left( \frac{1}{6} \right)^k

for

k = 1, 2, 3, \ldots

Explanation

P(N = k) = \frac{1}{6} \left( \frac{5}{6} \right)^{k-1}

for

k = 1, 2, 3, \ldots

Explanation

P(N = k) = \frac{5}{6} \left( \frac{1}{6} \right)^{k-1}

for

k = 1, 2, 3, \ldots

Explanation

P(N = k) = \frac{1}{6} \left( \frac{1}{6} \right)^{k-1}

for

k = 1, 2, 3, \ldots

Explanation

2. A person 'S' is driving to work every working day. The driving time of 'S' is between 7 to 10 minutes on a sunny day and 30 to 45 minutes on a rainy day, with all times being equally likely in each case. In a month of 30 days, 3 weeks are said to be sunny while the rest are rainy. What is the PDF of the driving time as a random variable of X?

f_X(x) = \frac{7}{30}, \text{ for } 7 \leq x \leq 10

and

f_X(x) = \frac{3}{150}, \text{ for } 30 \leq x \leq 45

, and

f_X(x) = 0 \text{ otherwise}

Explanation

f_X(x) = \frac{1}{30}, \text{ for } 7 \leq x \leq 10

and

f_X(x) = \frac{2}{150}, \text{ for } 30 \leq x \leq 45

, and

f_X(x) = 0 \text{ otherwise}

Explanation

f_X(x) = \frac{21}{30}, \text{ for } 7 \leq x \leq 10

and

f_X(x) = \frac{9}{150}, \text{ for } 30 \leq x \leq 45

, and

f_X(x) = 0 \text{ otherwise}

Explanation

f_X(x) = \frac{9}{30}, \text{ for } 7 \leq x \leq 10

and

f_X(x) = \frac{3}{150}, \text{ for } 30 \leq x \leq 45

, and

f_X(x) = 0 \text{ otherwise}

Explanation

3. The time until a small meteorite first lands anywhere in a desert is modeled as an exponential random variable with a mean of 10 days. The time is currently midnight. What is the probability that the meteorite first lands between 8 am on the 4th day and 12 pm on the 7th day?

P\left( \frac{10}{3} \leq X \leq \frac{13}{2} \right) = e^{-\frac{1}{3}} - e^{-\frac{13}{20}}

Explanation

P\left( \frac{10}{3} \leq X \leq \frac{13}{2} \right) = 1 - e^{-\frac{1}{3}} - e^{-\frac{13}{20}}

Explanation

P\left( \frac{10}{3} \leq X \leq \frac{13}{2} \right) = e^{-\frac{1}{3}} - e^{-\frac{13}{20}} - 1

Explanation

P\left( \frac{10}{3} \leq X \leq \frac{13}{2} \right) = e^{-\frac{1}{3}} + e^{-\frac{13}{20}}

Explanation

4. You take an exam that contains 10 MCQs. Each question has 4 possible options. You know the answers to the first 5 questions and for the remaining 5, you have randomly guessed an answer. Assume each question gives a score of 1 when your answer is correct and 0 for an incorrect one. Your score X on the exam is the total number of correct answers. What is

P(X > 8)

P(X > 8) = \frac{9}{2^{10}}

Explanation

P(X > 8) = \frac{1}{4^5}

Explanation

P(X > 8) = \frac{5}{2^{10}}

Explanation

P(X > 8) = \frac{10}{2^9}

Explanation

5. The number of customers arriving at a grocery store is modeled as a Poisson random variable. On average, 10 customers arrive in an hour. Let X be the number of customers arriving from 10:30 am to 11:30 am. What is

P(10 < X \leq 15)

P(10 < X \leq 15) = \sum_{k=11}^{15} \frac{e^{-15} \cdot 15^k}{k!}

Explanation

P(10 < X \leq 15) = \sum_{k=10}^{15} \frac{e^{-15} \cdot 15^k}{k!}

Explanation

P(10 < X \leq 15) = \sum_{k=11}^{15} \frac{e^{-10} \cdot 10^k}{k!}

Explanation

P(10 < X \leq 15) = \sum_{k=10}^{15} \frac{e^{-10} \cdot 10^k}{k!}

Explanation