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

Question

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

Bunuel · Accepted Answer

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.

KarishmaB · Answer

Responding to a pm:

The point is that the probability of picking a red will not depend on which draw it is.

In the first draw, probability of picking a red is 2/8 = 1/4

Probability of picking a red in the second draw:

2 cases:

Case 1: First draw is non red.
Probability = (6/8)*(2/7)
This is equal to (2/8)(6/7). Think about it: Probability of picking non red first and then red will be the same as probability of picking a red first and then a non red.

Case 2: First draw is red.
Probability = (2/8)*(1/7)
This is the probability of picking a red first and then a red again.

Total probability of second draw being red = (2/8)*(6/7) + (2/8)*(1/7) = 2/8
This is just the probability of picking a red first and then any ball (non red or red). Probability of picking ANY ball will be 1. Hence, the probability of picking a red in the second draw will be the same as the probability of picking a red in the first draw.

Similarly probability of picking a red in any draw will be the same.

anshumishra · Answer

Favorable outcomes :

WWR -> P(WWR)= 3/8 * 2/7 * 2/6 = 12/336
WRR -> P(WRR)= 3/8 * 2/7 * 1/6 = 6/336
RWR -> P(RWR)= 2/8 * 3/7 * 1/6 = 6/336
BBR -> P(BBR)= 3/8 * 2/7 * 2/6 = 12/336
BRR -> P(BRR)= 3/8 * 2/7 * 1/7 = 6/336
RBR -> P(RBR)= 2/8 * 3/7 * 1/6 = 6/336
WBR -> P(WBR)= 3/8 * 3/7 * 2/6 = 18/336
BWR -> P(BWR)= 3/8 * 3/7 * 2/6 = 18/336

So, P(Total) = 84/336 = 1/4 = 0.25

mainhoon · Answer

In problems such as these, the answer is just the probability based on the initial population... so 2/8 and it would be the same answer regardless of the nth try, 1st, 5th, 8th...

anshumishra · Answer

Excellent Bunuel! 
I have seen a couple of such nice, handful and timesavers scattered across different posts . Have you considered posting a sticky where you collect them at one place. 
That would be very helpful , also the GMatClub may add it to the math section of the IPhone app.

Thanks

https://gmatclub.com/static/posted_from.png     Posted from    GMAT ToolKit

zisis · Answer

!!!!!
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}

psychomath · Answer

P(drawing a red ball) = 1-P(not drawing a red ball )
=> 1-6C1/8C1 = 1/4

zisis · Answer

IMO when someone masters the ability to use combinations and probability at the same time in a gmat question, he/she has reached the level to move to a different topic....

@psychomath nice solution...

imhimanshu · Answer

Hi Karishma,

Thank You for the explanation.

Just curious to know, do we have any scenario where this phenomenon fails.

Thanks

maaadhu · Answer

Bunuel,

Why is it not 
7c2 * 1 / 8c3 = 3/8?

selecting any 2 without 1 red is 7c2....selecting the last red is 1...so fav outcomes = 7c2 * 1

total outcomes = 8c3

prob = 7c2/8c3

Where am I going wrong?

please explain..

Bunuel · Answer

Don't quite understand what you are doing there. What does "selecting any 2 without 1 red is 7c2" mean? Anyway, the question basically asks: if we put all 8 balls in a row, what is the probability that the third ball is red? Similar questions might help: a-box-contains-3-yellow-balls-and-5-black-balls-one-by-one-90272.html a-bag-contains-3-white-balls-3-black-balls-2-red-balls-100023.html each-of-four-different-locks-has-a-matching-key-the-keys-101553.html if-40-people-get-the-chance-to-pick-a-card-from-a-canister-97015.html new-set-of-mixed-questions-150204-100.html#p1208473 a-medical-researcher-must-choose-one-of-14-patients-to-127396.html

maaadhu · Answer

Bunuel,

first 2 balls can be selected from any of   - so selecting any 2 balls from 7 balls is 7c2.

probability of selecting the remaining red ball is 1.

so fav outcomes = 7c2*1.

probability = 7c2/8c3 = 3/8/

I know my thinking is definitely wrong.

Can you please point out the defect in my thinking?

KarishmaB · Answer

Responding to a pm:

There are a lot of problems here.

If you want to use combinations here, you can do this:

Assuming all balls are distinct, Number of ways of selecting 3 balls one after another without replacement = 7*6*1*2
(Keep one red ball away for the third pick. This can be done in 2 ways. Now of the 7 remaining balls, pick 1 for the first pick and then another for the second pick.)

Total number of outcomes = 8*7*6

Probability = (7*6*2)/(8*7*6) = 1/4

nguyenduong · Answer

The big mistake is here that you ignored the phrase of "drawing without replacement", then your probability didn't exclude 1 ball picked up previously.

nguyenduong · Answer

The big mistake is here that you ignored the phrase of "drawing without replacement", then your probability didn't exclude 1 ball picked up previously.

Asifpirlo · Answer

the probability of the 3rd ball not to be Red = 6/8
so, the required probability = 1- 6/8 = 2/8 = 0.25

jlgdr · Answer

Not quite sure because after 2 draws there won't be 8 balls anymore no?
Also, what happens if we draw a red in the first two? Then won't the probability of drawing a red in the third draw be zero?

Cheers
J

Engineer1 · Answer

By "In problems such as these", you meant "Replacement Problems", right?
Just want to be sure so that I can apply it for similar problems. : )

Bunuel · Answer

Problems such as these: a-box-contains-3-yellow-balls-and-5-black-balls-one-by-one-90272.html a-bag-contains-3-white-balls-3-black-balls-2-red-balls-100023.html each-of-four-different-locks-has-a-matching-key-the-keys-101553.html if-40-people-get-the-chance-to-pick-a-card-from-a-canister-97015.html new-set-of-mixed-questions-150204-100.html#p1208473 a-medical-researcher-must-choose-one-of-14-patients-to-127396.html

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

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