Question

A bowl contains 8 blue marbles and 5 green marbles. Elizabeth randomly draws 4 marbles from the bowl. She does not replace the marbles after each draw. What is the probability that she draws 1 blue marble and 3 green marbles?
A.
B.
C.
D.

Answer 1

Probability ( 1 blue marble and 3 green marbles) = ( Blue, green, green, green) + ( green, blue, green, green) + ( green, green, blue, green) + ( green, green, green, blue) = { (8/13) × (5/12) × (4/11) × (3/10) } + { (5/13) × (8/12) × (4/11) × (3/10) } + { (5/13) × (4/12) × (8/11) × (3/10) } + { (5/13) × (4/12) × (3/11) × (8/10) } = (4/143) + (4/143) + (4/143)+ (4/143) = 16/143 If the answer does not correspond to yours, do tell me. I'll try to correct mine

Answer 2

P (1 blue, 3 green) = $\frac{C (8, 1)*C(5, 3)}{C(13, 4)} = \frac{80}{715}= \frac{16}{143} }$