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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This note was uploaded on 02/09/2014 for the course ISYE 2027 taught by Professor Zahrn during the Spring '08 term at Georgia Tech.
 Spring '08
 Zahrn
 The Land

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