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Bunuel
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IMO D
Brendon. scoring equal to or more than 700: b = 4/10, Brendon. scoring less than 700: not b = 6/10
Daniel scoring equal to or more than 700: k = 5/10, Daniel. scoring less than 700: not k = 5/10
Kane scoring equal to or more than 700: d = 6/10, Daniel. scoring less than 700: not d = 4/10

P(at least 2 of them score less than 700) = P(b * not k * not d) + P(not b * k * not d) + P(not b * not k * d) + P(not b * not k * not d)
= (4/10 * 5/10* 4/10) + (6/10 * 5/10 * 4/10) + (6/10 * 5/0* 6/10) + (6/10* 5/10* 4/10)
= 80/1000 + 120/1000 + 180/1000 + 120/1000 = 500/1000 = 1/2
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Hey Bunuel, what is the correct method to solve this question?
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Bunuel
If the probability that Brendon, Daniel and Kane score more than or equal to 700 on the GMAT is 0.4, 0.5 and 0.6 respectively, what is the probability that at least 2 of them score less than 700?

A. 0.12
B. 0.38
C. 0.40
D. 0.50
E. 0.60

Solution:

We see that the probability that Brendon, Daniel, and Kane score less than 700 on the GMAT is 0.6, 0.5, and 0.4 respectively. The probability that at least 2 of them score less than 700 is the sum of the probability that exactly 2 of them score less than 700 and the probability that all 3 of them score less than 700. Therefore, we have:

There are 3 ways in which exactly 2 score less than 700. Letting Y = Yes and N = No, the 3 ways are (YYN), (YNY), and (NYY). We calculate the probability of each and add the three probabilities.

P(exactly 2 score less than 700) = 0.6 x 0.5 x 0.6 + 0.6 x 0.5 x 0.4 + 0.4 x 0.5 x 0.4 = 0.18 + 0.12 + 0.08 = 0.38

(Note: In the above calculation, the bolded factors are the probabilities that they score less than 700 and the unbolded factors are the probabilities that they score greater than or equal to 700.)

There is only 1 way in which all of them score less than 700: (YYY).

P(all 3 score less than 700) = 0.6 x 0.5 x 0.4 = 0.12

Therefore, P(at least 2 of them score less than 700) = 0.38 + 0.12 = 0.50.

Answer: D
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