# Probability that the first randomly picked ball is white.

Another mystery for me:

There are 4 white balls and 7 black balls in the jar. Two balls are taken from the jar randomly and not put back in the jar. The second(last) ball is white. What is the probability that the first ball was also white?

Now I think that it shouldn't matter that the second ball was white because the second pick can't affect the first one.

so I think the probability should be 4/11 = 0.36 ...

Am I correct or am I wrong?

• Are you familiar with Bayes' Law and the concept of conditional probability? If the balls were replaced then it is true that knowing the result of the second pick would not affect your state of knowledge of the first, as the events of the first and second pick are assumed independent of each other in this case. You are told, however, that the balls have not been replaced and so the result of the second pick is not independent of the first. Commented Sep 26, 2015 at 21:42
• @JWeissman I think the question is misleading me.We know for sure that the second picked ball is white. Now the question is what are the chances that the first picked ball was also white. So it is referring to the event that has happened in the past. As I'm picking randomly the FIRST ball from the jar I do not care about the future events. So I thought 4/11 is enough in this case. Commented Sep 26, 2015 at 21:55
• If you think “it shouldn’t matter that the second ball was white, because the second pick can't affect the first one,” consider this question: There are 3 black balls and 1 white ball in a jar. Two balls are chosen randomly and not put back. The second ball is white. What is the probability that the first ball was also white? Commented Sep 26, 2015 at 22:02

No need to use Bayes Rule/conditional probabilities.

Pick the second ball first (when you pick a pair of balls, you can swap balls/change their order and preserve probabilities). You are left with 3 whites out of total 10 balls to draw a white ball from. The needed probability is 3/10

• Simplest approach (+1) Commented Mar 15, 2016 at 7:02

Let $P(W_1)$ be the probability that the first ball is white, and $P(W_2)$ the probability that the second ball is white.

$$P(W_1\cap W_2)=\frac 4{11}\frac 3{10}=\frac 6{55}\\ p(W_2)=\frac 4{11}\frac 3{10}+\frac7{11}\frac 4{10}=\frac{20}{55}$$

So, $$P(W_1|W_2)=\frac{P(W_1\cap W_2)}{P(W_2)}=\frac 6{20}=.3$$

• Thank you! Nicely explained. Commented Sep 26, 2015 at 21:58
• Your quite welcome. :) Commented Sep 26, 2015 at 21:59

The probability of drawing W-W is (4/11)*(3/10)

The probability of drawing B-W is (7/11)*(4/10)

So the probability of the first being white given that the 2nd was white is

(4/11)(3/10) / ( (4/11)(3/10) + (7/11)*(4/10) ) which reduces to 3/10

• Hmm, nice, never thought I should calculate it like this. Thank you! Commented Sep 26, 2015 at 21:56