Rich has 3 green, 2 red and 3 blue balls in a bag. He randomly picks 5 from the bag WITH replacement. What is the probability that of the 5 drawn balls, Rich has picked 1 red, 2 green, and 2 blue balls?

GGBBR prob = 3/8*3/8*3/8*3/8*2/8=3^4*2/8^5
GGBBR can have 3C2*3C2*2C1 combination = 3*3*2=18

Total prob = 18*3^4*2/8^5

i think damager is absolutely right. we have replacements, so the total number of combinations is 8^5 (n*n*n*n*n) and not 8c5...what's OA?

your reasoning sounds better...what's OA?

For me the probability is:

P=[5!/(2!·2!)]*[(3/8)^2]*[(3/8)^2]*[(2/8)^1]=2430/16384=0.15

What is the OA?

I think the answer should be 9/28. Permutation assumes replacement. Without replacement would get really tricky.

P=2/8*(3/8*2/8)*(3/8*2/8)

Let's clear out this one:

CASE I.
Rich has 3 green, 2 red and 3 blue balls in a bag. He randomly picks 5 from the bag WITH replacement. What is the probability that of the 5 drawn balls, Rich has picked 1 red, 2 green, and 2 blue balls?

The probability would be: \frac{2}{8}*(\frac{3}{8}*\frac{3}{8})*(\frac{3}{8}*\frac{3}{8})*\frac{5!}{2!2!}=14.8%

CASE II.
If the problem was WITHOUT replacement:

Then the probability would be: \frac{2}{8}*(\frac{3}{7}*\frac{2}{6})*(\frac{3}{5}*\frac{2}{4})*\frac{5!}{2!*2!}=\frac{9}{28}=32.14%

OR different way of counting: \frac{2C1*3C2*3C2}{8C5}=\frac{9}{28}


The only difference we have that in FIRST case we are always drawing from 8 balls and the initial number of balls in bag remains the same, as we are putting withdrawn balls back (we have replacement).

In the SECOND case after each withdrawal we are reducing total number of balls by 1 AND reducing the withdrawn balls number by 1.

We are then multiplying this, in BOTH CASES, by \frac{5!}{2!2!}, because we want to calculate number of ways this combination is possible (1 red, 2 green, 2 blue), total combination of 5 balls is 5!, BUT as among these 5 balls we have 2 green and 2 red, we should divide 5! by 2!2! to get rid of repeated combinations.

The second way of calculating "WITHOUT replacement case" just counts the number of combinations of picking desired number of each ball from its total number (1 red out of two, 2 green out of three, and 2 blue out of three), multiplies them and divides this by the number of ways we can pick total of 5 balls out of 8.

Hope now it's clear.
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Thank you all, guys!
Understood it completely!

GMAT TIGER, order does not matter.
Because we have the following combination:

GGBBR

if we mark numbers G1G2B1B2R and rearrange them to have G2G1B2B1R, we will still have the same combination.
thus we eliminate repeats when we divide 5! (GGBBR) by 2! (BB) and 2! (GG)
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I think the ans is (3/8)^2*2/8*8!/(3!*3!*2!)