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# A bag contains 3 white balls, 3 black balls & 2 red balls

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Director
Joined: 23 Jan 2013
Posts: 560
Schools: Cambridge'16
Re: A bag contains 3 white balls, 3 black balls & 2 red balls  [#permalink]

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02 Aug 2015, 04:22
First option (first two balls are not red and the last is red): 6/8*5/7*2/6=5/28

Second option (first or second ball is red and the last is red): (6/8*2/7*1/6)*2=2/28

So, 5/28+2/28=7/28=1/4

A
Intern
Joined: 07 Jan 2016
Posts: 23
Schools: AGSM '18
Re: A bag contains 3 white balls, 3 black balls & 2 red balls  [#permalink]

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13 Mar 2016, 06:02
Could you tell me more about your way to give that formula? Thank you.

zisis wrote:
Bunuel wrote:
krishan wrote:
A bag contains 3 white balls, 3 black balls & 2 red balls. One by one three balls are drawn out without replacement. What is the probability that the third ball is red?

A. 0.25
B. 0.15
C. 0.35
D. 0.45
E. 0.40

Thanks

The initial probability of drawing red ball is 2/8. Without knowing the other results, the probability of drawing red ball will not change for ANY successive drawing: second, third, fourth...

Similar problems:
probability-of-picking-last-97015.html?hilit=initial%20probability

Hope it's clear.

!!!!!
went the long way - still under 2min

RedOtherRed +OtherRR+OOR

$$\frac{1}{28}+\frac{1}{28}+\frac{5}{28}=\frac{7}{28}=\frac{1}{4}$$
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Joined: 09 Sep 2013
Posts: 9461
Re: A bag contains 3 white balls, 3 black balls & 2 red balls  [#permalink]

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24 May 2018, 16:00
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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Re: A bag contains 3 white balls, 3 black balls & 2 red balls &nbs [#permalink] 24 May 2018, 16:00

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