are disjoint events p e1 e2 b p e1 b p

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Unformatted text preview: es that different numbers show on the dice decreases from the original probability that different numbers show on the dice. Example 2.3.2 Continuing with the previous example, find and compare P (B c ) and P (B c |A). Solution: Here, B c is the event that the numbers coming up are the same. Since six outcomes 1 6 are in B c , P (B c ) = 36 = 1 . Since AB c = {(3, 3)}, P (AB c ) = 36 . As noted above, P (A) = 5/36. 6 c) ( 1/36 Therefore, P (B c |A) = PPAB) = 5/36 = 1/5. Another way to compute P (B c |A) would have been to (A notice that P (B |A)+ P (B c |A) = 1, and use the fact, from Example 2.3.1, that P (B |A) = 4/5. Note that, for this example, P (B c |A) > P (B c ). That is, if one learns that the sum is six, the chances that the numbers coming up are the same number increases from the original probability that the numbers coming up are the same. Example 2.3.3 Again, consider the rolls of two fair dice. Let E =“the number showing on the first die is even,” and F =“the sum of the numbers showing is seven.” Find and compare P (F ) and P (F |E ). 6 1 Solution: Since F has six outcomes, P (F ) =...
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