Conditional part is chances of Ben losing.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
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-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.
This looks good to me.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 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.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
