carlyp28
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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.)

Respuesta :

mreijepatmat
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