Rich has 3 green, 2 red and 3 blue balls in a bag. He

(3C2*3C2*2C1)/8C5
= (3*3*2)/56=18/56=9/28

1. find combinations for each color separately:

green: xxo -> 3c2=3
blue: xxo -> 3c2=3
red: xo -> 2c1=2

2. find total combinations for picked balls:

xxxxxooo -> 5c8=56

3. find prob (selected over total cobmibations ):
P=3*3*2 / 56 = 9/28

B.

would the answer be the same with replacement?

I think it will be different.

Separate comb-s w/ replacemnt:
green: xxo -> 3c2=3*2*1=6
blue: xxo -> 3c2=3*2*1=6
red: xo -> 2c1=2*1=2

Total comb-s w/ replacement:
5c8 -> 8*7*6*5*4

Probability:
(6*6*2) / (8*7*6*5*4) = 3/280

ok guys !!! I am totally confused here about .With Replacement and Without Replacemetn Funda

..I thought With out replacement means we are taking out the balls and not putting it back.

so Total no of ways = 8*7*6*5*4

the no of ways the 1 Red can be drawn = 2
2 Green = 3*2
2 blue = 3*2

so prob = 6*6*2/8*7*6*5*4= 3/280.....


Can any body pls explain the funda of With replacement and Without Replacement.

The event happens without replacement. Could you please explain how that factors into this answer? Maybe the answer for with replacement?

Let's look at pick two green balls from three green balls. How many ways can you do it? Follow your train of thought, the first pick there's three possibility. The second pick there's only two left, so two possibility. Then you multiply them. However, notice that if the first pick you picked A, and then the second pick you picked B, you would count it as one outcome. If you picked B first, and then A second, you would count it as another outcome. While in fact, (A B) and (B A) is the same outcome, you picked the same two balls. So what you need to do in that case is that you need to divede your number by 2! to get rid of the ordering.

By the same logic for total outcome you have to divide 8*7*6*5*4 by 5!, and you'll get the same result as C(8,5).

Generally, picking n things one by one without replacement can be treated the same as picking n things all together.

Can somebody can put more light on this solution by probablit method with out using Combinations.

Thanks,
Rrsnathan.

The number of ways to select 1 red ball is 2C1 = 2.

The number of ways to select 2 green balls is 3C2 = 3.

The number of ways to select 2 blue balls is 3C2 = 3.

So, the number of ways to select 1 red, 2 green, and 2 blue balls is 2 x 3 x 3 = 18.

The number of ways to select 5 balls from 8 is 8C5:

8!/5![(8-5)!] = 8!/(5!x3!) = (8 x 7 x 6 x 5 x 4)/5! = (8 x 7 x 6 x 5 x 4)/(5 x 4 x 3 x 2 x 1) = 7 x 2 x 4 = 56

Thus, the probability is 18/56 = 9/28.

Answer: B

\(3G,2R,3B\,\,\,\, \to \,\,{\text{take}}\,\,5\,\,{\text{of}}\,\,{\text{them}}\,\,{\text{at}}\,{\text{random}}\)

\({\text{? = P}}\left( {1\,\,{\text{red}},\,\,2\,\,{\text{green}},\,\,2\,\,{\text{blue}}} \right) = \frac{{C\left( {2,1} \right) \cdot C\left( {3,2} \right) \cdot C\left( {3,2} \right)}}{{C\left( {8,5} \right)}} = \frac{{2 \cdot 3 \cdot 3}}{{8 \cdot 7}} = \frac{9}{{28}}\)

This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.

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