udaymathapati wrote:

A box contains red and blue balls only. If there are 8 balls in total, how many red balls are in the box?

(1) If two balls are randomly selected without replacement, the probability that both balls are red is 5/14

(2) If two balls are randomly selected without replacement, the probability is 15/56 that the first ball is red and the second ball is blue.

Please explain your appraoch

Lets say there are R red balls and 8-R blue balls.

S1: P(Both-Red) = \(\frac{R}{8}*\frac{R-1}{7} = \frac{5}{14}\)

so R(R-1) = 5*4, hence R = 5 (considering only positive value).

SufficientS2: P(Red-Blue) = \(\frac{R}{8}*\frac{8-R}{7} = \frac{15}{56}\)

so R(8-R) = 15, hence R = 5 or 3.

Not SufficientAnswer: A

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Respect,

Vaibhav

PS: Correct me if I am wrong.