What is the most efficient way to achieve the result for question attached?
Answer: E
Thanks,
K
Answer: E
Thanks,
K
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P(White or Even) = P(White) + P(Even) - P(White and Even)kwah wrote:What is the most efficient way to achieve the result for question attached?
Answer: E
Thanks,
K
(1) Probability that the ball will both be white and have an even number painted on it is 0 implies P(White and Even) = 0
So, P(White or Even) = P(White) + P(Even) - 0, which is clearly NOT sufficient.
(2) Probability that the ball will be white minus the probability that the ball will have an even number painted on it is 0.2 implies P(White) - P(Even) = 0.2, which is again NOT sufficient as there can be a number of possibilities.
Combining (1) and (2), P(White and Even) = 0 and P(White) - P(Even) = 0.2
P(White or Even) = 0.2 + P(Even) + P(Even) - 0
= 2P(Even) + 0.2, which is NOT sufficient as there can be a number of possibilities.
The correct answer is E.
