Problem Solving

A jar contains 6 Magenta balls, 3 Tan balls, 5 Gray balls and 7 Turquoise balls. Two balls are chosen from the jar. What is the probability that both balls chosen are Tan?

1) 1/70
2) 2/49
3) 1/21
4) 6/441
5) 1/49

Don't we have to assume that the balls are identical? If they are, then the number of ways of choosing 2 Tan identical balls out of 3 tan balls = 1 (since they are identical, order doesn't matter). Isn't it? ---- My answer was C) Is it a trap? Really confused
If I assume that the balls are not withdrawn "AT A TIME", OA is correct. However, how do I know whether the balls are withdrawn at a time or one by one?

OA A

Source : Veritas

Brent@GMATPrepNow wrote:

voodoo_child wrote:A jar contains 6 Magenta balls, 3 Tan balls, 5 Gray balls and 7 Turquoise balls. Two balls are chosen from the jar. What is the probability that both balls chosen are Tan?

1) 1/70
2) 2/49
3) 1/21
4) 6/441
5) 1/49

P(both tan) = P(1st ball is tan AND 2nd ball is tan)
= P(1st ball is tan) X P(2nd ball is tan)
= (3/21) X (2/20)
= 1/70 = A

It doesn't matter if the two balls are selected at once or one at a time (as long as there's no replacement). When I say "1st ball is tan," I just mean the first ball we examine.

Cheers,
Brent

Thanks Brent - I have a question. Isn't it that there is only one way to choose 2 balls out of 3 identical balls? We can't use 3C2 (as you did above) because all the elements are same. Let's say there are three letters A A A - how many ways can I choose 2 As? Just one because ALL the A's are the same. IT doesn't matter which A is chosen. Isn't it?

Can you please explain what I am missing here?

Thanks Brent. That sounds correct to me because there are 1000 black balls available. If the question were - choose two black balls, would we apply 1000c2? I don't think so. Not sure.

However, now I am a bit confused by your previous post on https://www.beatthegmat.com/combinations ... 15538.html

In this post, we concluded that we can use nCr to choose r elements from n non-identical balls. The poster asked you whether we can pick Green balls in (2C0=1)+(2C1=2)+(2C2=1)= 4 ways. I guess this works only when balls are not identical. If the balls were identical, the answer would be 1 (choose 1 ball) +1 (choose 2 balls) +1(choose 3 balls) = Isn't it?

I am a bit confused. Please help me...

I got it. Please ignore....

3С2/21С2=1/70