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Conditional Probability

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by satishchandra » Tue Nov 22, 2011 9:48 pm
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) 49/72
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by Anurag@Gurome » Tue Nov 22, 2011 9:57 pm
satishchandra 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) 49/72
Conditional part is chances of Ben losing.

Ben losing and then Rob winning = 6/7 * 1/3 = 6/21 = 2/7

Ben losing and then Mike winning = 6/7 * 1/4 = 6/28 = 3/14

2/7 + 3/14 = 4/14 + 3/14 = 7/14 = 1/2

The correct answer is C.
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by satishchandra » Tue Nov 22, 2011 10:53 pm
Anurag@Gurome wrote: Conditional part is chances of Ben losing.

Ben losing and then Rob winning = 6/7 * 1/3 = 6/21 = 2/7

Ben losing and then Mike winning = 6/7 * 1/4 = 6/28 = 3/14

2/7 + 3/14 = 4/14 + 3/14 = 7/14 = 1/2

The correct answer is C.
I am not sure about the formulation of the question itself. I would agree and do the same steps as Anurag, were the question saying-
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, if Ben were to lose the championship?
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by GmatMathPro » Sat Nov 26, 2011 8:45 am
I think the wording of the question is fine. For Mike or Rob to win the championship, Ben has to lose. If Ben doesn't lose (that is, he wins), then Mike and Rob each have a 0% chance of winning the championship. The only, very slight, assumption that we are making is that exactly one person can win the championship. But that is a very reasonable assumption.

Your rephrasing changes the nature of the question. For one thing, with your new wording, the probabilities of Mike or Rob winning are not conditional upon Ben losing. The way you have it, it sounds like 1/4 and 1/3 is the chance of either Rob or Mike winning before anything happens with Ben. Now, if you're saying that Ben is definitely losing the championship, this fact presumably impacts the probabilities of Mike and Rob winning.

There is a lengthier discussion about this problem here: https://www.beatthegmat.com/need-expert- ... 92238.html
Last edited by GmatMathPro on Sat Dec 03, 2011 6:57 am, edited 1 time in total.
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by satishchandra » Sun Nov 27, 2011 1:40 am
I read through your link. I think the basic flaw was in my understanding the problem itself. Now, I changed my approach, writing the formulae, which I learnt in school. I am not quite sure whether the formulae I wrote are right; However, they are yielding to a right answer.

P(B)= Ben wins
P(B!)= Ben loses
P(M)= Mike Wins
P(M/B!) = Mike wins conditional on Bob losing = 1/4
P(R/B!) = Rob wins conditional on Bob losing = 1/3

P(M U R) = P(M) + P(R) - P(M ∩ R)
= P(M/B!)*P(B!) + P(R/B!)*P(B!) - 0
= 6/7*1/4 + 6/7*1/3 = 1/2

Are the approach and the formulae I wrote correct??


Now using the same approach as per my wording of the question,

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, if Ben were to lose the championship?

P(M)= 1/4
P(R) = 1/3
P(B) = 1/7

P(M/B! U R/B!) = P(M)/P(B!)+ P(R)/P(B!) - 0
=(1/4)/6/7) + (1/3)/(6/7) = 7/6(7/12) = 49/72


MathPro,
What do you think about my approaches? Any flaw in them?
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by amit2k9 » Mon Nov 28, 2011 11:22 pm
(6/7)*(1/3+1/4)
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by GmatMathPro » Sat Dec 03, 2011 8:30 am
satishchandra wrote:I read through your link. I think the basic flaw was in my understanding the problem itself. Now, I changed my approach, writing the formulae, which I learnt in school. I am not quite sure whether the formulae I wrote are right; However, they are yielding to a right answer.

P(B)= Ben wins
P(B!)= Ben loses
P(M)= Mike Wins
P(M/B!) = Mike wins conditional on Bob losing = 1/4
P(R/B!) = Rob wins conditional on Bob losing = 1/3

P(M U R) = P(M) + P(R) - P(M ∩ R)
= P(M/B!)*P(B!) + P(R/B!)*P(B!) - 0
= 6/7*1/4 + 6/7*1/3 = 1/2

Are the approach and the formulae I wrote correct??
This looks good to me.
Now using the same approach as per my wording of the question,

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, if Ben were to lose the championship?

P(M)= 1/4
P(R) = 1/3
P(B) = 1/7

P(M/B! U R/B!) = P(M)/P(B!)+ P(R)/P(B!) - 0
=(1/4)/6/7) + (1/3)/(6/7) = 7/6(7/12) = 49/72
This looks okay too, as long as we assume that the relative magnitudes of the probabilities are the same as they were before we found out that Ben lost. Absent any information to the contrary, this is a reasonable assumption to make on an artificial problem such as this one, but be aware that it is still an assumption. But again, this is a correct application of the concept of conditional probability if we make this assumption.
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