Related questions with answers

You have a six-sided die that you roll once and observe the number of dots facing upwards. What is the sample space? What is the probability of each sample outcome? What is the probability of E, the event that the roll is even?

Computer programs are classified by the length of the source code and by the execution time. Programs with more than 150 lines in the source code are big (B). Programs with $\leq$ 150 lines are little (L). Fast programs (F) run in less than 0.1 seconds. Slow programs (W) require at least 0.1 seconds. Monitor a program executed by a computer. Observe the length of the source code and the run time. The probability model for this experiment contains the following information: P [LF]=0.5, P[BF]=0.2, and P[BW]=0.2. What is the sample space of the experiment? Calculate the following probabilities: P[W], P[B], and $\mathrm{P}[W \cup B]$ .

You roll a 6-sided die repeatedly. Starting with roll i=1, let $R_{i}$ denote the result of roll i. If $R_{i}>i$ , then you will roll again; otherwise you stop. Let N denote the number of rolls. (a) What is P[N>3]? (b) Find the PMF of N.

Question

You have a six-sided die that you roll once. Let $R_{i}$ denote the event that the roll is i. Let $G_{j}$ denote the event that the roll is greater than j. Let E denote the event that the roll of the die is even-numbered. (a) What is $\mathrm{P}\left[R_{3} | G_{1}\right]$ , the conditional probability that 3 is rolled given that the roll is greater than 1? (b) What is the conditional probability that 6 is rolled given that the roll is greater than 3? (c) What is $\mathrm{P}\left[G_{3} | E\right]$ , the conditional probability that the roll is greater than 3 given that the roll is even? (d) Given that the roll is greater than 3, what is the conditional probability that the roll is even?

Solution

Verified

Step 1

1 of 3

Sample Space for rolling a die, S = {1, 2, 3, 4, 5, 6}

Since, all the outcomes are equally likely, P(rolling any number) = $\dfrac{1}{6}$

$\textbf{(a)}$

As we can see, $G_{1}$ is the sample space in this case.

$G_{1}$ = {2, 3, 4, 5, 6}

Hence, P( $G_{1}$ ) = P( $R_{2}$ ) + P( $R_{3}$ ) + P( $R_{4}$ ) + P( $R_{5}$ ) + P( $R_{6}$ )

Hence, P( $G_{1}$ ) = $\dfrac{1}{6} + \dfrac{1}{6} + \dfrac{1}{6} + \dfrac{1}{6} + \dfrac{1}{6} = \dfrac{5}{6}$

and $R_{3}$ = {3}

Hence, $G_{1} \cap R_{3}$ = {3} = $R_{3}$

Hence, P( $G_{1} \cap R_{3}$ ) = P( $R_{3}$ ) = $\dfrac{1}{6}$

Hence, the conditional probability that 3 is rolled given that the roll is greater than 1 i.e.

P( $R_{3} | G_{1}$ ) = $\dfrac{P(G_{1} \cap R_{3})}{P(G_{1})}$ = $\dfrac{\frac{1}{6}}{\frac{5}{6}}$ = $\boxed{0.2}$

$\textbf{(b)}$

As we can see, $G_{3}$ is the sample space in this case.

$G_{3}$ = {4, 5, 6}

Hence, P( $G_{3}$ ) = P( $R_{4}$ ) + P( $R_{5}$ ) + P( $R_{6}$ )

Hence, P( $G_{3}$ ) = $\dfrac{1}{6} + \dfrac{1}{6} + \dfrac{1}{6} = \dfrac{3}{6}$

and $R_{6}$ = {6}

Hence, $G_{3} \cap R_{6}$ = {6} = $R_{6}$

Hence, P( $G_{3} \cap R_{6}$ ) = P( $R_{6}$ ) = $\dfrac{1}{3}$

Hence, the conditional probability that 6 is rolled given that the roll is greater than 3 i.e.

P( $R_{6} | G_{3}$ ) = $\dfrac{P(G_{3} \cap R_{6})}{P(G_{3})}$ = $\dfrac{\frac{1}{6}}{\frac{3}{6}}$ = $\boxed{0.33}$

Create an account to view solutions

By signing up, you accept Quizlet's Terms of Service and Privacy Policy

Continue with Google Continue with Facebook

Create an account to view solutions

By signing up, you accept Quizlet's Terms of Service and Privacy Policy

Continue with Google Continue with Facebook

Related questions with answers

Create an account to view solutions

Create an account to view solutions

Recommended textbook solutions

Probability and Stochastic Processes: A Friendly Introduction for Electrical and Computer Engineers

Probability Concepts in Engineering: Emphasis on Applications to Civil and Environmental Engineering

Probability and Stochastic Processes: A Friendly Introduction for Electrical and Computer Engineers

Probability and Stochastic Processes

More related questions