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.
OA A
Source: Magoosh
A box contains red and blue balls only. If there are 8 balls
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We have to determine the number of red balls.BTGmoderatorDC 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.
OA A
Source: Magoosh
Say the number of red balls is r; thus, the number of blue balls = 8 - r.
Let's take each statement one by one.
(1) If two balls are randomly selected without replacement, the probability that both balls are red is 5/14
=> rC2 / 8C2 = r*(r - 1)/8.7 = 5/14 => r*(r - 1) = 20 => r*(r - 1) = 5*4 => r = 5. Sufficient.
(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.
Probability of drawing the first ball being red = r/8;
Probability of drawing the second ball being blue = (8 - r) / 7; there would be only 7 balls left after the drawing the first ball.
Thus, the required probability = r/8 * (8 - r) / 7 = 5/14
=> r*(8 - r) = 15
r*(8 - r) = 3*5
r = 3 or 5. No unique answer. Insufficient.
The correct answer: A
Hope this helps!
-Jay
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