A box contains 6 black balls and 4 white balls. If two balls are selec

Question

A box contains 6 black balls and 4 white balls. If two balls are selected at random without replacement, what is the probability that both balls are white?

A. 7/90
B. 3/25
C. 2/15
D. 4/25
E. 4/9

yenbh · Answer

The probability of both balls are white = 2C4/ 2C10 = 2/15

Hence, choice C is the answer

Princ · Answer

OA:C

Probability that both ball are white = \frac{C(4,2)}{C(10,2)}= \frac{4!}{2!2!}*\frac{2!8!}{10!}=\frac{4*3}{2*1}*\frac{2*1}{10*9}=\frac{2}{15}

ScottTargetTestPrep · Answer

There are 10 balls in the box. The probability that the first ball drawn will be white is 4/10, and the probability that the second ball will also be white is 3/9. Thus, the probability of selecting two white balls, without replacement, is 4/10 x 3/9 = 2/5 x 1/3 = 2/15.

Answer: C

gmat1393 · Answer

Ways of selecting 2 white balls = 4C2
probability of selecting 2 white balls= 4C2/10C2 =2/15

Option C is the answer.

michael30 · Answer

1st pick
P(1st white ball) =  \frac{4}{10}

2nd pick (since without replacement, there are only 3 white balls and 9 total number of balls)
P(2nd white ball) =  \frac{3}{9}

P(2 white balls) =  \frac{4}{10}  *  \frac{3}{9}  =  \frac{2}{15}

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