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30 September, 19:50

Assume that two fair dice are rolled. First compute P (F) and then P (F|E). Explain why one would expect the probability of F to change as it did when we added the condition that E had occurred.

F: the total is two

E: an even

total shows on the dice

Compute P (F).

P (F) equals=

nothing

(Simplify your answer.)

+5
Answers (1)
  1. 30 September, 22:30
    0
    Sample space = {6 x 6} = 36 outcomes

    Only rolling "1" and "1" gives a sum of 2 (one favorable outcome)

    (F) = P (1+1=2) = 1/36

    P (F/E) = P (F∩E) / P (E)

    If E = even and already occurred we can't have F = 2, since E = 2 or 4 or 6

    Then in this case P (F/E) = 0 since P (F∩E) = 0. and P (F) = 0
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