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probability doubt 15 Mar 2010, 23:58
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Dear Gurus,

I came across this question from another board.

A manager of Team in Pool A is analyzing his teams winning chances for a big game. The probability of his Team winning against is – 30% against Half of the other teams; 50% against Quarter of the other teams; 40% against the last Quarter remaining;
What is the probability of his teams winning if the opponent is picked at random?
(A) 0.175
(B) 0.5
(C) 0.375
(D) 0.3
(E) 0.25

The other board didnt publish any answers for this yet. Any pointers please?

Peace n Love,
Max..



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probability doubt 16 Mar 2010, 07:21
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C

Basically you multiply the probability of choosing a specific group (the half, the quarter, or the other quarter) by the probability of winning against a team in that group. You then add all 3 products.

(1/2 * 30) + (1/4 * 50) + (1/4 * 40) = 150/4 = 37.5 (you dont even have to calculate the final answer, you know 150/4 is less than 50 and more than 30 thus C is the only possible answer).

Note that we assume that the groups don't over lap (i.e. part of the 'half' group is in one of the 'quarter' groups), but I guess thats implied by the question.
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probability doubt 16 Mar 2010, 07:57
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nickk
C

Basically you multiply the probability of choosing a specific group (the half, the quarter, or the other quarter) by the probability of winning against a team in that group. You then add all 3 products.

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Can you explain above statement?
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probability doubt 16 Mar 2010, 08:05
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sidhu4u
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C

Basically you multiply the probability of choosing a specific group (the half, the quarter, or the other quarter) by the probability of winning against a team in that group. You then add all 3 products.

Can you explain above statement?
Show more

I think the best would be to draw a tree diagram but I'll try to explain verbally.
The 'average' probability of winning (so if you pick one opponent at random) is the weighted average probability of winning against each. So in our case, we can imagine that we only have 3 opponents, A, B, and C. The weight we assign to each is the probability of choosing that opponent at random. In our case, A: 1/2 (if we pick one team at random, there is a 50% chance that it will be from group A, which is the 50% group), B: 1/4, C: 1/4.
Next, we multiple the probabilities of winning against each opponent by the weight and sum.
A: 1/2 * 30% = 15%
B: 1/4 * 50% = 12.5%
C: 1/4 * 40% = 10%

The sum of the above products is 37.5% = 0.375
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probability doubt 19 Mar 2010, 12:59
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sidhu4u
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C

Basically you multiply the probability of choosing a specific group (the half, the quarter, or the other quarter) by the probability of winning against a team in that group. You then add all 3 products.

Can you explain above statement?

I think the best would be to draw a tree diagram but I'll try to explain verbally.
The 'average' probability of winning (so if you pick one opponent at random) is the weighted average probability of winning against each. So in our case, we can imagine that we only have 3 opponents, A, B, and C. The weight we assign to each is the probability of choosing that opponent at random. In our case, A: 1/2 (if we pick one team at random, there is a 50% chance that it will be from group A, which is the 50% group), B: 1/4, C: 1/4.
Next, we multiple the probabilities of winning against each opponent by the weight and sum.
A: 1/2 * 30% = 15%
B: 1/4 * 50% = 12.5%
C: 1/4 * 40% = 10%

The sum of the above products is 37.5% = 0.375
Show more


Thanks a lot...got it now. :)



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