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by hey_thr67 » Tue Jun 26, 2012 10:53 am
If Ben were to lose the championship, Mike would be the winner with a probability of 1/4, and Rob 1/3. If the probability of Ben being the winner is 1/7, what is the probability that either Mike or Rob will win the championship?

A: 1/12
B: 1/7
C: 1/2
D: 7/12
E: 6/7
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by eagleeye » Tue Jun 26, 2012 2:14 pm
This is a combined probability problem.

Let Probability of Ben, Mike and Rob winning be P(B), P(M), P(R) respectively.

Then we are told that, when Ben doesn't win, the probability of Mike winning is 1/4 and that of Rob is 1/3. Now probability of Ben losing = P(#B) = 1-P(B). (where I have used #B to show the event of Ben not winning. Remember that probability of one winning and the other two also winning is 0.
The event that Ben loses and Mike wins is denoted by (#BM). Now P(#BM) = P(#B)*P(M).

So, we are given that
P(#B)*P(M) = 1/4
P(#B)*P(R) = 1/3
Then we are told that P(B) = 1/7. So P(#B) = 1-1/7 = 6/7.

We need to find probability of either Mike or Rob winning. So we need P(M)+P(R).

In the first two equations, substituting P(#B) = 6/7 and adding them, we get
6/7*(P(M)+P(R)) = 1/4 + 1/3 => P(M)+ P(R) = (1/3+1/4)*6/7 = 7/12*6/7 = 1/2.

Hence C is the right answer.

Let me know if this helps :)
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by Mike@Magoosh » Tue Jun 26, 2012 2:21 pm
hey_thr67 wrote:If Ben were to lose the championship, Mike would be the winner with a probability of 1/4, and Rob 1/3. If the probability of Ben being the winner is 1/7, what is the probability that either Mike or Rob will win the championship?

A: 1/12
B: 1/7
C: 1/2
D: 7/12
E: 6/7
Hi, there. I'm happy to help with this. :-)

This is a tricky question. First of all, P(Ben wins) = 1/7, so P(Ben doesn't win) = 1 - 1/7 = 6/7.

The wording of the first part is very important: If Ben were to lose the championship, Mike would be the winner with a probability of 1/4, and Rob 1/3. This is a conditional probability --- we are told the probability given that a certain condition (Ben losing) is met.

The conditional probability of A, given B, is written P(A|B). The formula for this is

P(A and B) = P(B)*P(A|B)

Here, B = Ben loses, and we want A = Mike or Rob wins.

P(Mike wins|Ben loses) = 1/4
P(Rob wins|Ben loses) = 1/3

Add these two:
P(Mike wins|Ben loses) + P(Rob wins|Ben loses) = P(Mike or Rob wins|Ben loses) = 1/4 + 1/3 = 7/12

P(Mike or Rob wins) = P(Ben loses)*P(Mike or Rob wins|Ben loses) = 6/7 * 7/12 = 6/12 = 1/2

Answer = C

Again, that was a very hard question. This is at the very outer limit of what the GMAT could possible ask. Here's a much more typical kind of GMAT math probability question.

https://gmat.magoosh.com/questions/1036
When you submit your answer to this question, the next page will have the full video solution.

Let me know if you have any more questions.

Mike :-)
Magoosh GMAT Instructor
https://gmat.magoosh.com/
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by GMATGuruNY » Tue Jun 26, 2012 3:21 pm
hey_thr67 wrote:If Ben were to lose the championship, Mike would be the winner with a probability of 1/4, and Rob 1/3. If the probability of Ben being the winner is 1/7, what is the probability that either Mike or Rob will win the championship?

A: 1/12
B: 1/7
C: 1/2
D: 7/12
E: 6/7
When we want one event AND another event to happen -- P(A AND B) -- we MULTIPLY the fractions.
When we want one event OR another event to happen -- P(A OR B) -- we ADD the fractions.

Since P(Ben wins) = 1/7, P(Ben loses) = 6/7.

Case 1: Ben loses AND Mike wins.
Since we want both events to happen, we multiply:
6/7 * 1/4 = 3/14.

Case 2: Ben loses AND Rob wins.
Since we want both events to happen, we multiply:
6/7 * 1/3 = 2/7.

Since either Case 1 OR Case 2 will yield a good outcome, we add the results above:
3/14 + 2/7 = 3/14 + 4/14 = 7/14 = 1/2.

The correct answer is C.
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