Given an urn containing 3 red balls and 4 white balls. If two balls are drawn and the first ball is not replaced before the second is drawn, what is the probability that both balls are red?
A) 2/7
B) 7/12
C) 1/6
D) 1/7
E) 1/5

Question

Quizplus · Accepted Answer

When the first ball is drawn, there are 3 red balls and 4 white balls in the urn. Therefore, the probability of drawing a red ball on the first draw is 3/7 and the probability of drawing a red ball on the second draw (without replacement) is 2/6 or 1/3, since there are now 2 red balls and 6 total balls left in the urn. To find the probability of both events occurring, we multiply the probabilities: 
P(both balls are red) = (3/7) × (1/3) = 1/7
Therefore, the correct answer is D.

Given an Urn Containing 3 Red Balls and 4 White