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13 Jun 2019, 00:32
Kudos
Bookmarks
C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
75%
(hard)
Question Stats:
62% (02:54) correct
38%
(03:01)
wrong
based on 182
sessions
History
Date
Time
Result
Not Attempted Yet
A certain sports league has ten teams in its Western Conference and eight teams in its Eastern Conference. At the end of the season, two teams, one from each conference, play in The Big Game. Two teams in each conference are located in Texas. If each team in each conference has an equal probability of making it to The Big Game, and if each team in The Big Game has an equal probability of winning that game, what is the probability that a team from Texas will win The Big Game?
A. 3/20
B. 2/9
C. 9/40
D. 27/80
E. 5/16
A. 3/20
B. 2/9
C. 9/40
D. 27/80
E. 5/16
04 Jul 2019, 10:06
Kudos
Bookmarks
10 Western Conference Teams of which 2 are from Texas = 1/5 chance a Texas team makes the big game
8 Eastern Conference Teams of which 2 are from Texas = 1/4 chance a Texas team makes the big game
Now the Western probability of (1/5) needs to be multiplied by the 50% chance of winning in finals so (1/5)*(1/2) = 1/10
The Eastern Probability of (1/4) needs to be multiplied by the 50% chance of winning in finals so (1/4)*(1.2) = 1/8
So now that you have each conferences probability, you need to add them. Because its either a eastern conference team wins OR a western conference team wins so (1/10)+(1/8) = 9/40
8 Eastern Conference Teams of which 2 are from Texas = 1/4 chance a Texas team makes the big game
Now the Western probability of (1/5) needs to be multiplied by the 50% chance of winning in finals so (1/5)*(1/2) = 1/10
The Eastern Probability of (1/4) needs to be multiplied by the 50% chance of winning in finals so (1/4)*(1.2) = 1/8
So now that you have each conferences probability, you need to add them. Because its either a eastern conference team wins OR a western conference team wins so (1/10)+(1/8) = 9/40
General Discussion
13 Jun 2019, 00:47
Kudos
Bookmarks
Bunuel
Either both Texas teams can be selected or only 1 Texas team can be selected in the Big game for a Texas team to win
Probability that both Texas teams are selected from each conference = 2/10*2/8 = 4/80
--> Probability that a Texas team wins = 4/80*1 (1 --> both teams are from Texas) = 4/80
Probability that only 1 Texas team is selected
= selected from western or selected from eastern
= 2/10*6/8 + 8/10*2/8
= 28/80
--> Probability that a Texas team wins = 28/80*1/2 (1/2 --> s only 1 team is from Texas) = 14/80
So, total probability = 4/80 + 14/80 = 18/80 = 9/40
IMO Option C
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