Got Homework?
Connect with other students for help. It's a free community.
Here's the question you clicked on:
 0 viewing
Two fair dice are rolled. What is the conditional probability that at least one lands on 6 given that the dice land on different numbers. Can someone please take a look at my work and show me where I am going wrong?
 2 years ago
 2 years ago
Two fair dice are rolled. What is the conditional probability that at least one lands on 6 given that the dice land on different numbers. Can someone please take a look at my work and show me where I am going wrong?
 2 years ago
 2 years ago

This Question is Closed

bahrom7893Best ResponseYou've already chosen the best response.1
okay, so this is what I have: P(A)  probability that at least 1 is 6 P(B)  probability that the die land on two diff numbers
 2 years ago

bahrom7893Best ResponseYou've already chosen the best response.1
There are the following options for at least one die to be 6: {(1,6);(2,6),(3,6),(4,6),(5,6),(6,6),(6,5),(6,4),(6,3),(6,2)(6,1)}
 2 years ago

bahrom7893Best ResponseYou've already chosen the best response.1
P(B) = 1  P(both land on the same number) = 1  (6/36) = 1(1/6) = 5/6
 2 years ago

bahrom7893Best ResponseYou've already chosen the best response.1
since there are 6 ways in which we can have two same numbers.
 2 years ago

bahrom7893Best ResponseYou've already chosen the best response.1
P(AB) = P(A U B)/P(B)
 2 years ago

bahrom7893Best ResponseYou've already chosen the best response.1
P(A U B) is they are different and at least one die is 6, so we just take the options above: {(1,6);(2,6),(3,6),(4,6),(5,6),{{(6,6)}},(6,5),(6,4),(6,3),(6,2)(6,1)}, but take away (6,6) So P(A U B) = 10/36
 2 years ago

bahrom7893Best ResponseYou've already chosen the best response.1
So now we have: P(AB) = (10/36)/(5/6) = (10/36)*(6/5) = (2/6) = 1/3
 2 years ago

bahrom7893Best ResponseYou've already chosen the best response.1
OMG.. i got it right lol
 2 years ago

bahrom7893Best ResponseYou've already chosen the best response.1
sorry for bothering you guys, but i was being dumb earlier
 2 years ago

ZarkonBest ResponseYou've already chosen the best response.1
your notation is not correct...should be... \[P(AB)=\frac{P(A\cap B)}{P(B)}\]
 2 years ago

bahrom7893Best ResponseYou've already chosen the best response.1
yea P(AintersectB), not P(AUB)
 2 years ago
See more questions >>>
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.