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

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Director
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G
Joined: 23 Jan 2013
Posts: 597
Schools: Cambridge'16
Re: A bag contains 3 white balls, 3 black balls & 2 red balls [#permalink]

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New post 02 Aug 2015, 05:22
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
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Re: A bag contains 3 white balls, 3 black balls & 2 red balls [#permalink]

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New post 13 Mar 2016, 07:02
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:
ps-probability-gmat-made-me-lose-my-marbles-36830.html?hilit=initial%20draw
probability-of-picking-last-97015.html?hilit=initial%20probability
probability-question-please-solve-90272.html?hilit=initial%20probability#p686028

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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Re: A bag contains 3 white balls, 3 black balls & 2 red balls [#permalink]

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New post 24 May 2018, 17:00
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Re: A bag contains 3 white balls, 3 black balls & 2 red balls   [#permalink] 24 May 2018, 17:00
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