imhimanshu wrote:
There are 10 solid colored balls in a box, including 1 Green and 1 Yellow. If 3 of the balls in the box are chosen at random, without replacement, what is the probability that the 3 balls chosen will include the Green ball but not the yellow ball.
A. 1/6
B. 7/30
C. 1/4
D. 3/10
E. 4/15
The number of ways to select 3 balls that include the green ball but not the yellow ball is:
1C1 x 1C0 x 8C2 = 1 x 1 x (8 x 7)/2! = 28 ways
(Note: 1C1 is the number of ways to select 1 green ball from the 1 green ball, 1C0 is the number of ways to select 0 yellow ball from the 1 yellow ball, and 8C2 is the number of ways to select 2 balls from the 8 remaining balls.)
The number of ways to select 3 balls from 10 is:
10C3 = (10 x 9 x 8)/3! = (10 x 9 x 8)/(3 x 2) = 120
Thus, the probability that the 3 balls chosen will include the Green ball but not the yellow ball is 28/120 = 7/30.
Alternate Solution:
Let’s first determine the probability of selecting the one green ball on the first draw and any of the remaining balls (except the yellow ball) on the succeeding two draws: 1/10 x 8/9 x 7/8 = 56/720.
Now, we see that we could also pick the green ball on the second draw, with any ball (except the yellow one) on the first and third draws, with probability: 8/10 x 1/9 x ⅞ = 56/720.
Similarly, we could instead pick the green ball on the third draw, with any ball (except the yellow one) on the first and second draws, with probability: 8/10 x 7/9 x 1/8 = 56/720.
Thus the total probability for the event where the green ball is drawn but the yellow one is not is:
56/720 + 56/720 + 56/720 = 168/720 = 21/90 = 7/30.
Answer: B
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