Question

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?

Answer 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

Answer 2

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)}

Answer 3

So: P(A) = 11/36

Answer 4

P(B) = 1 - P(both land on the same number) = 1 - (6/36) = 1-(1/6) = 5/6

Answer 5

since there are 6 ways in which we can have two same numbers.

Answer 6

P(A|B) = P(A U B)/P(B)

Answer 7

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

Answer 8

So now we have:
P(A|B) = (10/36)/(5/6) = (10/36)*(6/5) = (2/6) = 1/3

Answer 9

OMG.. i got it right lol

Answer 10

sorry for bothering you guys, but i was being dumb earlier

Answer 11

haha ok

Answer 12

your notation is not correct...should be...

\[P(A|B)=\frac{P(A\cap B)}{P(B)}\]

Answer 13

yea P(AintersectB), not P(AUB)