It is currently 16 Dec 2017, 16:41

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

# A certain sports league has ten teams in its Western Confere

Author Message
TAGS:

### Hide Tags

Manager
Joined: 16 Feb 2011
Posts: 195

Kudos [?]: 247 [1], given: 78

Schools: ABCD
A certain sports league has ten teams in its Western Confere [#permalink]

### Show Tags

30 Sep 2012, 17:10
1
KUDOS
2
This post was
BOOKMARKED
00:00

Difficulty:

55% (hard)

Question Stats:

67% (01:33) correct 33% (02:13) wrong based on 24 sessions

### HideShow timer Statistics

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

What would be the probability of P{East winning the Super bowl}. Is it 1/2? Or 8/18 because of equally likely probabilities?

Just curious. Here's why I am asking:

Here's what I did initially:
P(T=texas wins) = P(Texas Wins/East) P(East) + P (Texas Wins/West) P(West)
= 10/18 * 2/10 + 8/18 * 2/8
= 2/9

If I change P(Texas Wins/East) = P(Texas Wins/East) = 1/2, ( the question says that "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," --- I am not sure whether this is an accurate translation...)

then P(texas wins) = 1/2 * 2/10 + 1/2 * 2/8 = 9/40... Any thoughts? Please help....
[Reveal] Spoiler: OA

Kudos [?]: 247 [1], given: 78

Senior Manager
Joined: 28 Aug 2006
Posts: 303

Kudos [?]: 174 [1], given: 0

Re: A certain sports league has ten teams in its Western Confere [#permalink]

### Show Tags

30 Sep 2012, 22:08
1
KUDOS
voodoochild 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

What would be the probability of P{East winning the Super bowl}. Is it 1/2? Or 8/18 because of equally likely probabilities?

Just curious. Here's why I am asking:

Here's what I did initially:
P(T=texas wins) = P(Texas Wins/East) P(East) + P (Texas Wins/West) P(West)
= 10/18 * 2/10 + 8/18 * 2/8
= 2/9

If I change P(Texas Wins/East) = P(Texas Wins/East) = 1/2, ( the question says that "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," --- I am not sure whether this is an accurate translation...)

then P(texas wins) = 1/2 * 2/10 + 1/2 * 2/8 = 9/40... Any thoughts? Please help....

Following are the three cases
1. Texas team from West, no Texas team from East and Texas team wins = 2/10 * 6/8 *1/2 = 3/40
2. No Texas team from West, Texas team from East and Texas team wins = 8/10 * 2/8 *1/2 = 4/40
3. Texas team from West, Texas team from East (since both are texas teams and one will win, this is certain event) = 2/10 * 2/8 = 2/40
Hence, the required probability = 3/40+4/40+2/40 = 9/40
_________________

Kudos [?]: 174 [1], given: 0

Manager
Joined: 16 Feb 2011
Posts: 195

Kudos [?]: 247 [0], given: 78

Schools: ABCD
Re: A certain sports league has ten teams in its Western Confere [#permalink]

### Show Tags

01 Oct 2012, 03:19
thanks. That's another way to solve the problem. Do you think that my solution is incorrect? Thoughts?

Kudos [?]: 247 [0], given: 78

Non-Human User
Joined: 09 Sep 2013
Posts: 14813

Kudos [?]: 288 [0], given: 0

Re: A certain sports league has ten teams in its Western Confere [#permalink]

### Show Tags

11 Jul 2017, 18:55
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________

Kudos [?]: 288 [0], given: 0

Re: A certain sports league has ten teams in its Western Confere   [#permalink] 11 Jul 2017, 18:55
Display posts from previous: Sort by