# of total marbles in the jar equals to 8+y, out of which R=8 and W=y

P(RR)=\frac{8}{8+y}*\frac{8-1}{8+y-1}=\frac{8}{8+y}*\frac{7}{7+y};
P(RW)=2*\frac{8}{8+y}*\frac{y}{8+y-1}=2*\frac{8}{8+y}*\frac{y}{7+y}, multiplying by 2 as RW can occure in two ways RW or WR;

Question: is \frac{8}{8+y}*\frac{7}{7+y}>2\frac{8}{8+y}*\frac{y}{7+y}? --> is \frac{7}{2}>y? --> is y<3.5? eg 0, 1, 2, 3.

(1) y ≤ 8, not sufficient.
(2) y ≥ 4, sufficient, (y is not less than 3.5).

Answer: B.
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For probability there is no difference whether we pick some # of marbles one by one or all at once.

For example:
Jar containing 3 red and 5 blue marbles. We pick 2 marbles one by one. What is the probability that both marbles are red?
Jar containing 3 red and 5 blue marbles. We pick 2 marbles simultaneously. What is the probability that both marbles are red?

\frac{C^2_3}{C^2_8}=\frac{3}{28} or \frac{3}{8}*\frac{2}{7}=\frac{3}{28}.

As for the "permutation": what do you mean by that? Question asks to compare the probability "of 2 red marbles" with the probability of "one marble of each color". So can you pleas clarify what permutation has to do with it?
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Bunuel, consider the case where Y = 2. then The probability of P(RW) = 2*\frac{8}{10}*\frac{2}{9} = \frac{32}{90}

P(RR) = \frac{8}{10}*\frac{7}{9} = \frac{56}{90}

So even for y=2, it is likely to pick two reds than a red and white. Isn't that contradicting the OA? As a matter of fact, as ykeeps growing this likelyhood goes down and so y >!8 (y cannot be greater than 8).

Or did I keep my foot in my mouth?

I think i didn't understand the problem properly. Thank you for explaining and sorry for the trouble.

I used combinatorics approach. Can someone clarify whether the approach is correct? Please.

Total = (8+y)C2
RR = 8C2
one from each = 8C1 * yC1

Probability of RR = \frac{8C2}{(8+y)C2}
Proabaility of 1 from each = \frac{8C1 * yC1}{(8+y)C2}

Question: \frac{8C2}{(8+y)C2} > \frac{8C1 * yC1}{(8+y)C2}?

Canceling denominator from both the sides : 8C2 > 8C1 * yC1 ?
\frac{(8 * 7)}{2} > 8 * ( \frac{y!}{(y-1)!}) ?
\frac{(8 * 7)}{2} > 8 * y ?

\frac{7}{2} > y ?

Thanks

(1) Prob 2 red marbles = 8/16 * 7/15 = 7/30

Prob of 2 marbles of diff color = 8/16 * 8/15 + 8/16 * 8/15 = 2 * 8/16 * 8/15 = 8/15 = 16/30

if y = 0, then we have a different result

So (1) is not sufficient


(2)

Prob 2 red marbles = 8/12 * 7/11 = 14/33

Prob of 2 marbles of diff color = 8/12 * 8/11 + 8/12 * 8/11 = 2 * 8/12 * 8/11 = 32/33

Anything greater than 4 also gives the same result.

So (2) is sufficient

Answer - B
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56/(8+y) * (7+y) > 16/(8+y) * (7+y)

56 > 16y => y > 3.

Clear B.
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Responding to a pm:
Order matters. That is the reason we multiply by 2 in the probability approach. We can pick R and then W or W and then R. So R and W can be picked in two different sequences with the same end result.

As for the combination approach - it works and you can use it but I wouldn't like to in this case. When we say 8C1, we are assuming that the balls are distinct and we can select one ball out of 8 Red balls in 8 ways. But it is kind of implied in the question that the balls are identical. Still, you can assume all the balls to be distinct and select 1 out of 8 Red balls in 8 ways and 1 out of y white balls in y ways.
Number of ways of selecting R and W = 8*y
Number of ways of selecting W and R = y*8
Total number of ways of selecting 2 balls = (8+y)(7+y) (first ball can be selected in 8+y ways and second in 8+y-1 ways)
Required probability = 16y/(8+y)(7+y) (which is the same as obtained in the probability method)
It works because in the numerator we assume all balls to be distinct and we do the same in the denominator too. Hence, the probability stays the same.

You can also calculate the probability assuming that order doesn't matter (for both numerator and denominator). Since it is probability, you will get the same probability since the 2 of the numerator will get canceled with the 2 of the denominator.

Number of ways of picking R and W = 8C1*yC1 (order doesn't matter so we don't multiply by 2)
Number of ways of picking 2 balls = (8+y)C2 (order doesn't matter when we use nCr)

Probability = 8C1*yC1/(8+y)C2 = 8y/(8+y)(7+y)/2 = 16y/(8+y)(7+y)
Obviously, the result is the same.
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You have four distinct balls. In how many ways can you select a ball? In 4 ways.
In how many ways can you select 2 balls? In 4C2 ways.
etc...
You have four identical balls. In how many ways can you select a ball? In 1 way. No matter which ball you pick, it is no different from the other 3 balls so every case is the same.
In how many ways can you select 2 balls? Again, only 1 way. You can select any two balls - they are all the same.

Check out the following posts for more on identical and distinct objects.
http://www.veritasprep.com/blog/2011/12 ... 93-part-1/
http://www.veritasprep.com/blog/2011/12 ... s-part-ii/
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