{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

tutorial2sol

# tutorial2sol - C wins the second game Solution Let E be the...

This preview shows pages 1–2. Sign up to view the full content.

Stat 211 - Tutorial 2 Solution/Answers 1. P ( A B ) = 0 . 1367 and P ( A B ) = 0 . 6133 . 2. P ( A B ) = 0 and P ( A B ) = 0 . 75 . 3. (a) 0.068608 (b) 0.140 4. Three people A , B and C are playing cards, each being equally likely to win the game. (a) In two consecutive games, i. find the probability that both A and B each win one game. Solution : Since each of A , B and C is equally likely to win, P ( A ) = P ( B ) = P ( C ) = 1 3 . Let A 1 be the event that A wins the 1st game, A 2 be the event that A wins the 2nd game, and so on. Tree diagram: First Game Second Game Sample Space Outcome C 1 C 2 C 1 C 2 P ( C 1 C 2 ) = 1 / 9 B 2 C 1 B 2 P ( C 1 B 2 ) = 1 / 9 A 2 C 1 A 2 P ( C 1 A 2 ) = 1 / 9 1/3 1/3 1/3 B 1 C 2 B 1 C 2 P ( B 1 C 2 ) = 1 / 9 B 2 B 1 B 2 P ( B 1 B 2 ) = 1 / 9 A 2 B 1 A 2 P ( B 1 A 2 ) = 1 / 9 1/3 1/3 1/3 A 1 C 2 A 1 C 2 P ( A 1 C 2 ) = 1 / 9 B 2 A 1 B 2 P ( A 2 B 2 ) = 1 / 9 A 2 A 1 A 2 P ( A 1 A 2 ) = 1 / 9 1/3 1/3 1/3 1/3 1/3 1/3 1

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
In the tree, the probability of A winning both games is given by P ( A 1 A 2 ) = P ( A 2 | A 1 ) P ( A 1 ) = (1 / 3)(1 / 3) = 1 / 9 . The other probabilities can be ob- tained in a similar manner. The event that both A and B each win one game is described by either the outcome A 1 B 2 or B 1 A 2 . Therefore, P ( both A and B each win one game ) = P ( A 1 B 2 ) + P ( B 1 A 2 ) = 1 / 9 + 1 / 9 = 2 / 9 ii. given that A wins at least one game, find the probability that
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: C wins the second game. Solution : Let E be the event that A wins at least one game. Then E = { A 1 A 2 ,A 1 B 2 ,A 1 C 2 ,B 1 A 2 ,C 1 A 2 } and so P ( E ) = P ( A 1 A 2 ) + P ( A 1 B 2 ) + P ( A 1 C 2 ) + P ( B 1 A 2 ) + P ( C 1 A 2 ) = 5 / 9 Therefore, the required probability is given by P ( C 2 | E ) = P ( C 2 ∩ E ) P ( E ) = P ( A 1 C 2 ) P ( E ) = 1 / 9 5 / 9 = 1 / 5 (b) Suppose instead that the probability that C wins a game is 5 / 7 , and that A and B are equally likely to win a game. Repeat part a). Solution : Here, P ( C ) = 5 / 7 , and P ( A ) = P ( B ) = 1 / 7 . Repeat part a) to obtain i. 0.0408 ii. 0.3846 5. (a) 0.56 (b) 0.857 6. Will be done in next week’s tutorial. 2...
View Full Document

{[ snackBarMessage ]}