slimsohn wrote:Hi, can someone help me with this problem- it seems simple, but I must be mistaken in my logic. Its from the Powerprep software Exam 2.

A certain basket contains 10 apples, 7 of which are red and 3 of which are green. If 3 different apples are to be selected at random from the basketbal, what is the probability that 2 of the apples selected will be red and 1 will be green?

A) 7/40
B) 7/20
C) 49/100
D) 21/40
E) 7/10

I thought it would just be 7/10 * 6/9 * 3/8, but thats wrong.
I'll post the answer after some discussion -- thanks.

Badri wrote:Assuming that the balls are not replaced in the basket after drawing:
There are three ways the incident can happen:


----1st Ball----2ndBall----3rd Ball---Probability
(1) Red--------Red--------Green-----P1 = ( 7/10 )*( 6/9 )*( 3/8 )
(2) Red--------Green------Red-------P2 = ( 7/10 )*( 3/9 )*( 6/8 )
(3) Green------Red--------Red-------P3 = ( 3/10 )*( 7/9 )*( 6/8 )


Total probability is P1 + P2 + P3
= 3*( 7*6*3 )/( 10*9*8 )
= 21/40

papgust wrote:I think we do not need to consider R1, R2 order here. Because a red ball is a red ball. You cannot distinguish 2 red balls.