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bahrom7893 Group Title

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

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  1. bahrom7893 Group Title
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    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
  2. bahrom7893 Group Title
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    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
  3. bahrom7893 Group Title
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    So: P(A) = 11/36

    • 2 years ago
  4. bahrom7893 Group Title
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    P(B) = 1 - P(both land on the same number) = 1 - (6/36) = 1-(1/6) = 5/6

    • 2 years ago
  5. bahrom7893 Group Title
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    since there are 6 ways in which we can have two same numbers.

    • 2 years ago
  6. bahrom7893 Group Title
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    P(A|B) = P(A U B)/P(B)

    • 2 years ago
  7. bahrom7893 Group Title
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    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
  8. bahrom7893 Group Title
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    So now we have: P(A|B) = (10/36)/(5/6) = (10/36)*(6/5) = (2/6) = 1/3

    • 2 years ago
  9. bahrom7893 Group Title
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    OMG.. i got it right lol

    • 2 years ago
  10. bahrom7893 Group Title
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    sorry for bothering you guys, but i was being dumb earlier

    • 2 years ago
  11. MelindaR Group Title
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    haha ok

    • 2 years ago
  12. Zarkon Group Title
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    your notation is not correct...should be... \[P(A|B)=\frac{P(A\cap B)}{P(B)}\]

    • 2 years ago
  13. bahrom7893 Group Title
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    yea P(AintersectB), not P(AUB)

    • 2 years ago
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