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.

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.

anonymous · Answer

a.13/143 b.4/13 c.4/143 d.16/143

kropot72 · Answer

$$\large P(1\ blue,\ 3\ green)=\frac{C(8, 1)\ \times C(5,\ 3)}{C(13,\ 4)}=\frac{8\times10}{715}$$

anonymous · Answer

i dont understand is that the answer? or how do you get the answer ?

kropot72 · Answer

No, it is part way to the answer. You need to simplify as follows:
|dw:1407006499399:dw|
Can you finish it now?