Isye 2027

# are disjoint events p e1 e2 b p e1 b p

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

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 ) =...
View Full Document

## 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.

Ask a homework question - tutors are online