bahrom7893
  • bahrom7893
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?
Mathematics
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SOLVED
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chestercat
  • chestercat
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bahrom7893
  • bahrom7893
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
bahrom7893
  • bahrom7893
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)}
bahrom7893
  • bahrom7893
So: P(A) = 11/36

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bahrom7893
  • bahrom7893
P(B) = 1 - P(both land on the same number) = 1 - (6/36) = 1-(1/6) = 5/6
bahrom7893
  • bahrom7893
since there are 6 ways in which we can have two same numbers.
bahrom7893
  • bahrom7893
P(A|B) = P(A U B)/P(B)
bahrom7893
  • bahrom7893
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
bahrom7893
  • bahrom7893
So now we have: P(A|B) = (10/36)/(5/6) = (10/36)*(6/5) = (2/6) = 1/3
bahrom7893
  • bahrom7893
OMG.. i got it right lol
bahrom7893
  • bahrom7893
sorry for bothering you guys, but i was being dumb earlier
anonymous
  • anonymous
haha ok
Zarkon
  • Zarkon
your notation is not correct...should be... \[P(A|B)=\frac{P(A\cap B)}{P(B)}\]
bahrom7893
  • bahrom7893
yea P(AintersectB), not P(AUB)

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