There are 10 solid-colored balls in a box, including 1 green ball and

Probability approach, direct not for the complementary event (it is more complicated, not worth trying in this case: it can be one G and also one Y or neither G nor Y, plus an additional different color):
P(G) * P(noG & noY) * P( another noG & noY) * 3 = 1/10 * 8/9 * 7/8 * 3 = 7/30.
We need the factor of 3 because the Green ball can be either the first, second or third.

Answer: B

Hi Bunuel,

Thanks for the reply. I was able to solve it using Approach # 2. But I have doubt in Approach # 1 . Request you to please provide your comments.

Isn't it an assumption that remaining 8 balls are of same color. If the remaining 8 balls each of them are of different color, then the number of arrangements will get changed and hence probability.

Please tell me what I am missing
Thanks.
H

We are not assuming that 8 other balls are necessarily of some one particular color: they can be of one color or 8 different colors, but for us the only thing which is important that they are of different color than green and yellow.

Consider this: if the first ball selected is green then we are left with 9 balls out of which 1 is yellow and 8 other balls are not yellow (that's the only thing we care), so the probability of selecting non-yellow ball is 8/9.

Hope it's clear.

Bunuel question for you--and it may be dumb. When I tried the problem I knew that there were three possibilities for getting the green ball. However, I ended up calculating the probabilities for each scenario. Do we know that the probably is the same for each scenario or could it be different?

The probability of each case will be the same:
{GXX} - \(P=\frac{1}{10}*\frac{8}{9}*\frac{7}{8}=\frac{7}{90}\);

{XGX} - \(P=\frac{8}{10}*\frac{1}{9}*\frac{7}{8}=\frac{7}{90}\);

{XXG} - \(P=\frac{8}{10}*\frac{7}{9}*\frac{1}{8}=\frac{7}{90}\);

The sum: \(\frac{7}{90}+\frac{7}{90}+\frac{7}{90}=\frac{7}{30}\).

Hope it helps.

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

Bunuel Need your help once more!

If we are choosing the balls without replacement, help me understand what is wrong with my approach:

3 possibilities: GXX, XGX, XXG

Probability = (1/10)(8/9)(7/8) + (9/10)(1/9)(7/8) + (9/10)(8/9)(1/8)
= (1/720)[ (56) + (63) + (72)]
= (1/720)[191]
= 191/720

Red figures are not correct.

Should be:
{GXX} - \(P=\frac{1}{10}*\frac{8}{9}*\frac{7}{8}=\frac{7}{90}\);

{XGX} - \(P=\frac{8}{10}*\frac{1}{9}*\frac{7}{8}=\frac{7}{90}\);

{XXG} - \(P=\frac{8}{10}*\frac{7}{9}*\frac{1}{8}=\frac{7}{90}\);

The sum: \(\frac{7}{90}+\frac{7}{90}+\frac{7}{90}=\frac{7}{30}\).

How are you able to do the frst approach calculation without calculator?

\(\frac{1}{10}*\frac{8}{9}*\frac{7}{8}*\frac{3!}{2!}=\frac{7}{30}\)

You don't need a calculator to get the value of the above. The key is to reduce.

8 gets reduced and we get:

\(\frac{1}{10}*\frac{1}{9}*7*\frac{3!}{2!}\)

\(\frac{3!}{2!}\) is just 3, so we get:

\(\frac{1}{10}*\frac{1}{9}*7*3\)

That 3 also gets reduced to give:

\(\frac{1}{10}*\frac{1}{3}*7\)

And finally, we arrive at:

\(\frac{7}{30}\)

Hope it helps.

Another tricky one!

We have three possible orders: 1) G/NY/NY 2) NY/G/NY, and 3) NY/NY/G. If we pick the "not yellow" balls before the green ball, then "not yellow" really means "not yellow OR green," because we have to save the green ball for later.

For example, in the order NY/NY/G, the first NY cannot be 9/10, because that includes the green ball, which we have to save for later. Thus, it must be 8/10 instead.

The same goes for the second NY, which must be 7/9.

Hi Bunuel

Could you explain why it's (1/10)*(8/9)*(7/8)*(3!/2!) and not (1/10)*(8/9)*(7/8)*(10!/7!3!) ? Thank you!

 

­3!/2! = 3 there represents the number of ways scenario (Green, Not Yellow, Not Yellow) can occur:

{Green, Not Yellow, Not Yellow}
{Not Yellow, Green, Not Yellow}
{Not Yellow, Not Yellow, Green}­

Thus, 3!/2! is essentially the number of permutations of GXX.