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Junior | Next Rank: 30 Posts
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by Bhandaripreeti » Mon Oct 15, 2007 5:05 am
What is the chance that Martha wins the first 5 rounds and loses the sixth?



(1) The chance that Martha wins the first 4 rounds and loses the fifth is 1/32



(2) The chance that Martha wins the first 6 rounds and loses the seventh is 1/128 .


Hi,

I thought the answer is D considering the numerator is 1 so the probability would be 1/2 however the OA is E .
Source - Princeton
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Source: — Data Sufficiency |
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by achandwa » Mon Oct 15, 2007 7:05 am
Let P(w1), P(w2), P(w3), P(w4), P(w5), P(w6), P(w7) denote the probably that Martha wins 1st, 2nd, 3rd, 4th, 5th, 6th, and 7th round respectively.
Let P(l1), P(l2), P(l3), P(l4), P(l5), P(l6), P(l7) denote the probably that Martha loses 1st, 2nd, 3rd, 4th, 5th, 6th, and 7th round respectively.

Probability that Martha wins the first 5 rounds and loses sixth = P(w1).P(w2).P(w3).P(w4).P(w5).P(l6)

(1) The chance that Martha wins the first 4 rounds and loses the fifth is 1/32
This provides you P(w1).P(w2).P(w3).P(w4).P(l5) - Insufficient to calculate the product above.

(2) The chance that Martha wins the first 6 rounds and loses the seventh is 1/128 .
This provides you P(w1).P(w2).P(w3).P(w4).P(w5).P(w6).P(l7) - Insufficient to calculate the product above.

Even taken together (1) and (2) are insufficient to calculate the produce P(w1).P(w2).P(w3).P(w4).P(w5).P(l6) - (E)
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