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

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.

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...

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

Posted from GMAT ToolKit

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

RedOtherRed +OtherRR+OOR

\(\frac{1}{28}+\frac{1}{28}+\frac{5}{28}=\frac{7}{28}=\frac{1}{4}\)

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

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...

Hi Karishma,

Thank You for the explanation.

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

Thanks

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..

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

Bunuel,

first 2 balls can be selected from any of [3 black + 3 white + 1 red (7 balls)] - 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?

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

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.

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.

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

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

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. : )

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