Problem Solving

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

Please explain your answer......

Here's what i did :
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
Apparently, it's incorrect.

Thanks

The probability of the team from Texas from western conf to make to the big game is 1/10
The probability of the team from Texas from eastern conf to make to the big game is 1/8
The probability of the team from Texas winning big game is 1/10 + 1/8 = [spoiler]9/40[/spoiler]

Guru's kindly correct me if i am wrong.

Interesting question - thanks for posting!

The reason "P(T=Texas wins) = P(Texas Wins/East) P(East) + P (Texas Wins/West) P(West)" does not work is because you do not need to multiply Texas probabilities with corresponding conference (East /West) probabilities - the Tx probability is included in the corresponding conference probability!

ntamhane has the right answer, I think (and so full points on the GMAT ), but the method is fuzzy...

"The probability of the team from Texas from western conf to make to the big game is 1/10"
Should be "The pr of the team from Tx from western conf to make to the big game is 2/10"
...and "The probability of the team from Texas from eastern conf to make to the big game is 1/8 "
should read: "The pr of the team from Tx from eastern conf to make to the big game is 2/8"

BECAUSE Tx has two teams in each conf, with equal likelihood to win!
("Two teams in each conference are located in Texas")

But as mentioned by ntamhane, I believe that the answer should be [spoiler]9/40[/spoiler]

Hi,
Probability that Texas team reaches final from West,p(w) = 2/10 =1/5
Probability that Texas team doesn't reach final from West,p(not w) = 1- 1/5 = 4/5
Probability that Texas team reaches final from East,P(e) = 2/8 = 1/4
Probability that Texas team doesn't reach final from East,p(not e) = 1- 1/4 = 3/4
Probability of a team reaching the Big game to win the final is 1/2
Probability that Texas team wins in the Big game is = p(w)*p(not e)*(1/2) + p(not w)*p(e)*(1/2) + p(e)*p(w)*1
=(1/5)(3/4)(1/2) + (4/5)(1/4)(1/2) + (1/5)(1/4)(1) = 9/40

Hence, C

probability of a team from Texas to reach finals in Western group = 2/10 = 1/5
probability of a team from Texas to reach finals in Eastern group = 2/8 = 1/4

probability of a Texan team to win from Western group = same for Eastern group = 1/2 each.

thus total probability = 1/5 * 1/2 + 1/4 * 1/2 = 9/40.
C. ( its an example of using both Independent events (product) and mutually exclusive events(sum)).

Hey Frank,

Thanks dude for the workout. Its great.

Frankenstein wrote:Hi,
Probability that Texas team ...... from East,p(not e) = 1- 1/4 = 3/4
Probability of a team reaching the Big game to win the final is 1/2
Probability that Texas team wins in the Big game is = p(w)*p(not e)*(1/2) + p(not w)*p(e)*(1/2) + p(e)*p(w)*1
=(1/5)(3/4)(1/2) + (4/5)(1/4)(1/2) + (1/5)(1/4)(1) = 9/40

Hence, C

Food for thought:

IF "Probability of a team reaching the Big game to win the final is 1/2"
And therefore: p(w)*p(not e) and p(not w)*p(e) have an equal chance of 1/2 each

Then what is the probability associated with p(e)*p(w)?
===========================

amit2k9's method is sound [final answer needs a minor edit(cannot be 9/20) though ] and here is some elaboration:

Pr | Event that East Conf Tx team goes to Big Game = 2/8 = 1/4

Pr | Event that West Conf Tx team goes to Big Game = 2/10 = 1/5

Pr | Event that East Conf Tx team wins Big Game = 1/2 of (Pr | Event that East Conf Tx team goes to Big Game) because both teams have an equal chance and ONLY one can win.

Similarly for West Conf also:
Pr | Event that W Conf Tx team wins Big Game = 1/2 of (Pr | Event that W Conf Tx team goes to Big Game) because both teams have an equal chance and ONLY one can win.

=> Pr | for each team must be multiplied by 1/2

(If you do not multiply by 1/2, then you are simply calculating the probability of the teams going to the Big Game, not winning)

=> Pr | team from Texas will win The Big Game = 1/2*1/4 + 1/2*1/5 = 1/8+1/10 = [spoiler]9/40[/spoiler]
Hope this helps!

Knewtonian wrote:

Frankenstein wrote:Hi,
Probability that Texas team ...... from East,p(not e) = 1- 1/4 = 3/4
Probability of a team reaching the Big game to win the final is 1/2
Probability that Texas team wins in the Big game is = p(w)*p(not e)*(1/2) + p(not w)*p(e)*(1/2) + p(e)*p(w)*1
=(1/5)(3/4)(1/2) + (4/5)(1/4)(1/2) + (1/5)(1/4)(1) = 9/40

Hence, C

Food for thought:

IF "Probability of a team reaching the Big game to win the final is 1/2"
And therefore: p(w)*p(not e) and p(not w)*p(e) have an equal chance of 1/2 each

Then what is the probability associated with p(e)*p(w)?

Hi,
When both Texas teams enter final, the probability that Texas team wins is 1.
Probability of both Texas teams entering final is p(e)*p(w).

Great thinking Frank!

So to sum this up, there are at least three logical solution structures:

1. (1/2)*(1/4) + (1/2)*(1/5) = 1/8+1/10 = 9/40

2. (1/5)(3/4)(1/2) + (4/5)(1/4)(1/2) + (1/5)(1/4)(1) = 9/40

3. 1*(1/4)*(1/5) + (1/2)*((1/4) - (1/4)*(1/5)) + (1/2)*(1/5-(1/4)*(1/5)) = 9/40
(# 3 anchors "(1/4)*(1/5)" to provide a single final step, albeit a long single step, solution)

Solution #1 is most parsimonious and recommended, #2 and #3 are similar in logic, correct and stylish but may take a few more seconds to solve - from a GMAT prep. perspective, it is best to understand all 3!

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.