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# A certain sports league has ten teams in its Western Conference and

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Math Expert
Joined: 02 Sep 2009
Posts: 55732
A certain sports league has ten teams in its Western Conference and  [#permalink]

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13 Jun 2019, 00:32
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(N/A)

Question Stats:

25% (04:05) correct 75% (03:22) wrong based on 4 sessions

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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

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Joined: 20 Jul 2017
Posts: 264
Re: A certain sports league has ten teams in its Western Conference and  [#permalink]

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13 Jun 2019, 00:47
Bunuel wrote:
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

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

Pls Hit Kudos if you like the solution
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Re: A certain sports league has ten teams in its Western Conference and  [#permalink]

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13 Jun 2019, 01:20
Probability that a team from Texas in the Western Conference reaches the Big Game, say P(TW) = 2/10 = 1/5
Probability that a team not from Texas in the Western Conference reaches the Big Game, (1-P(TW)) =1 - 2/10 = 8/10 = 4/5

Similarly,
Probability that a team from Texas in the Eastern Conference reaches the Big Game, P(TE) = 2/8= 1/4
Probability that a team not from Texas in the Western Conference reaches the Big Game, (1-P(TE) =1 - 2/8 = 6/8 = 3/4

If only one of the teams from Texas plays the Big Game, then the probability that the team from Texas win in that game is 1/2
If both teams in the Big Game are from Texas, we are not concerned about the outcome of the Big Game

Now, the probability that a team from Texas wins = P(TW)*(1-P(TE))*(1/2)+(1-P(TW))*P(TE)*(1/2)+P(TW)*P(TE)[/m]

$$=\frac{1}{5}*\frac{3}{4}*\frac{1}{2} + \frac{4}{5}*\frac{1}{4}*\frac{1}{2} + \frac{1}{5}*\frac{1}{4} = \frac{9}{40}$$

Hit Kudos if this helped!
Re: A certain sports league has ten teams in its Western Conference and   [#permalink] 13 Jun 2019, 01:20
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