Given a bag with 2 black balls and 3 white balls

Question

Given a bag with 2 black balls and 3 white balls, you randomly select one of them, in order. If the second ball you take is white, what is the probability that the first one was also white?

1. 2/3
2. 1/5
3. 3/4
4. 1/2
5. 3/5

Bunuel · Answer

This is a conditional probability question. Basically we are told that out of 2 balls picked one is white and are asked to find the probability of the other one also to be white.

So, possible options for this other ball would be: W, W, B, B (as one white is already picked). Hence the probability for that ball to be white too is 2/4=1/2.

Answer: D.

KarishmaB · Answer

Let me add a bit to what Bunuel said above:

Probability of event A = # of outcomes when A occurs/Total no of outcomes

When you have a condition given to you, the total no of outcomes adjust to that.

When you pick two balls, you can pick: {W, W}, {W, B}, {B, W} or {B, B}

But in this question, the total number of outcomes is not 4. We are given that the second ball is white. So the total number of outcomes is 2. Only {W, W} and {B, W} are possible.

What is the probability that the first one was also white? This happens in only 1 outcome - {W, W}

Probability = 1/2

Bunuel · Answer

Though this approach worked for this particular problem it might fail for others: suppose we have 10 white and 2 black balls and need to find the probability that the second ball is white while knowing that the first ball we picked was also white.

We also would have only two possible scenarios {W, W} and {W, B} but here we can not say that the answer is 1/2, because {W, W} and {W, B} will have the different # of ways to occur.

So, the answer for the above question would be 9/11 and not 1/2.

lpvinoth · Answer

Answer is 1/2.
Can anyone say which difficulty level this question would fall 500-600,600-700,700-750?

Bunuel · Answer

It's a 700 level question.

SVaidyaraman · Answer

Let w1, w2, w3, b1 and b2 be the balls.
1. The possible outcomes :
w1b1 
w1 b2
w2 b1
w2b2
w3 b1 
w3 b2
b1 w1
b2 w1
b1w2
b2 w2
b1w3
b2w3
w1 w2
w1 w3
w2 W3
w3 w1
w3 w2
w2 w1
b1 b2
b2 b1

But the total number of outcomes. is actually the number of times the second ball is white and the favorable outcomes is the number of times in that the first ball is also white.

2. The second ball is white on 12 occasions
3. The first ball is also white on 6 occasions
4. So the required probability is 6/12 = 1/2

Alternate approach:

1. we do not know what color is the first ball. Let it be of color x
2. The second ball is white
3. the 2 selections are x,W
4. Since we know the second ball is white, the first ball should have been filled by one among 2 white balls and 2 black balls
5. The probability of that being a white ball is 2/4=1/2

agourav · Answer

My approach is this:

Probability = No. of favorable cases/ total no. of cases

Total No. of cases = no. of times the second ball drawn is white.
No. of favorable cases = No. of cases in which the first ball drawn is also white given that the second ball drawn is white

Req. Probab. =  (Prob. of drawing a white ball in the first and second attempt)/{(Prob. of drawing a white ball in the first and second attempt)+(Prob. of drawing a black ball in the first attempt and a white ball in the second attempt)} 
Hence required probability = {(3/5)*(2/4)}/{(3/5)*(2/4) + (2/5)*(3/4)} = 1/2

Bunuel · Answer

And again, there is nothing wrong with the question.

bumpbot · Answer

Automated notice from GMAT Club BumpBot:

A member just gave Kudos to this thread, showing it’s still useful. I’ve bumped it to the top so more people can benefit. Feel free to add your own questions or solutions.

This post was generated automatically.

Given a bag with 2 black balls and 3 white balls

Prep Toolkit

Top 5 GMAT Debrief Videos

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards