Probability - balls - need expert help

Question

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If I assume that the balls are not withdrawn "AT A TIME", OA is correct. However, how do I know whether the balls are withdrawn at a time or one by one?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 · Answer

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

Brent Hanneson - Creator of GMATPrepNow.com

voodoo_child · Answer

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?

Brent@GMATPrepNow · Answer

Allow me to answer that question with another question:

A box contains 1000 identical black balls and 1 white ball. One ball is randomly selected.
What is P(black ball)?

The "identical" part has no bearing on the question. There are still 1001 outcomes and 1000 different ways to select a black ball.

P(black ball)=1000/1001

I hope that helps.

Cheers,
Brent

Brent Hanneson - Creator of GMATPrepNow.com

voodoo_child · Answer

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...

voodoo_child · Answer

I got it. Please ignore....

LalaB · Answer

3Ð¡2/21Ð¡2=1/70

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