A community for students. Sign up today!
Here's the question you clicked on:
 0 viewing

This Question is Closed

kropot72
 one year ago
Best ResponseYou've already chosen the best response.1The sample space has 36 possible combinations of numbers. These can be set out in column form as follows: 6,6 5,6 4,6 3,6 2,6 1,6 6,5 5,5 4,5 3,5 2,5 1,5 6,4 5,4 4,4 3,4 2,4 1,4 6,3 5,3 4,3 3,3 2,3 1,3 6,2 5,2 4,2 3,2 2,2 1,2 6,1 5,1 4,1 3,1 2,1 1,1 You need to count the number of outcomes where at least one die shows a four and divide the result by 36 to find the required probability.

kropot72
 one year ago
Best ResponseYou've already chosen the best response.1@mameadows Have you found the number of combinations where at least one die shows a four?

mameadows
 one year ago
Best ResponseYou've already chosen the best response.0well, I see from your answer that there are 11 possible combinations, but is there a formula that I can use to make this quicker to solve? so I don't always have to draw out all of the possible outcomes? thank you by the way... your answer really helped!

kropot72
 one year ago
Best ResponseYou've already chosen the best response.1No doubt there is an algebraic way to establish the number of combinations containing at least one four. However I think that making a table is the clearest way to solve this question.

kropot72
 one year ago
Best ResponseYou've already chosen the best response.1Here is another way to solve this question: Let event A be getting 4 on die #1 and 1, 2, 3, 5 or 6 on die #2. The probability of getting 4 on die #1 is 1/6 and the probability of getting 1, 2, 3, 5 or 6 on die #2 is 5/6. Therefore the probability of event A is \[P(A)=\frac{1}{6} \times \frac{5}{6}=\frac{5}{36}\] Let event B be getting 4 on die #2 and 1, 2, 3, 5 or 6 on die #1. The probability of getting 4 on die #2 is 1/6 and the probability of getting 1, 2, 3, 5 or 6 on die #1 is 5/6. Therefore the probability of event B is \[P(B)=\frac{5}{6} \times \frac{1}{6}=\frac{5}{36}\] Let event C be getting 4 on die #1 and 4 on die #2. The probability of event C is \[P(C)=\frac{1}{6} \times \frac{1}{6}=\frac{1}{36}\] Events A, B and C are mutually exclusive. Therefore the probability of getting at least one four is given by their sum, as follows: \[P(at\ least\ one\ 4)=\frac{5}{36}+\frac{5}{36}+\frac{1}{36}=\frac{11}{36}\]
Ask your own question
Ask a QuestionFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.