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Bob has a black box (you can't see what's inside the box). A long time ago Bob put one ball into the box but he doesn't remember what color the ball was. With equal probability it can be a white ball or a red ball. A. Bob takes a red ball and puts it in the same box. Now there are two balls in the box: one ball Red that Bob just put in and a ball that was in the box earlier (Bob doesn't remember its color). Now Bob draws Randomly one ball out of the box and it turned out to be a red ball. Calculate the probability that the ball that has been in the box for a long time is a white ball given the action taken by Bob.

My attempt: There are two options, since we already know of them is red, $A_1= \{\text{White, Red}\}$ or $A_2= \{\text{Red, Red}\}$, so $\Pr[A_1 \cup A_2] = 1/2$?

N. F. Taussig
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Mustafa Shama
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  • Bob put a red ball into the box then he took it back. Did it affect the probability? – Vasili Nov 21 '22 at 14:33
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    @Vasili The problem does not say that he took the same ball from the box, just that he extracted a red ball from the box. If there were two red balls in the box, he could have taken the one that was there before he added a red ball. – N. F. Taussig Nov 21 '22 at 15:01
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    Quick check why 1/2 is not the answer: if Bob would have drawn a white ball, the probability that the original ball was white would be 1, not 1/2. – svavil Nov 22 '22 at 02:31

4 Answers4

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You received two nice answers already, but let me add one that might be experienced as more intuitive.

If the ball has white color then the probability on drawing a red ball is $\frac12$.

If the ball has red color then the probability on drawing a red ball is $1$.

So comparing both situations you could say that the odds are $\frac12:1$.

We can also express that as $\frac13:\frac23$.

Here $\frac13+\frac23=1$ so that the two numbers can be interpreted as probabilities, and we can conclude that the first situation corresponds with probability $\frac13$.

drhab
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Two possibilities:

  • The first ball was white with probability $\frac12$. The conditional probability of then drawing a red ball would be $\frac12$ so the joint probability would be $\frac12 \times \frac12=\frac14$.
  • The first ball was red with probability $\frac12$. The conditional probability of then drawing a red ball would be $\frac22=1$ so the joint probability would be $\frac12 \times 1=\frac12$.

This means the marginal probability of drawing a red ball was $\frac14 +\frac12 = \frac34$.

So by Bayes' theorem, given a red ball was drawn, the posterior probability that the first ball was white (and similarly that the remaining ball is white) is $\dfrac{\frac14}{\frac34}=\dfrac13$.

Henry
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We are told that is it equally likely that Bob originally put a red ball in the box as it is that he put a white ball in the box. However, the additional piece of information that a red ball has been selected makes it less likely that the box contains a white ball.

Let $E$ be the event that the box contains a red ball and a white ball; let $F$ be the event that the box contains two red balls; let $R$ be the event that a red ball is selected from the box. Then we wish to calculate \begin{align*} \Pr(E \mid R) & = \frac{\Pr(E \cap R)}{\Pr(R)}\\ & = \frac{\Pr(E)\Pr(R \mid E)}{\Pr(E)\Pr(R \mid E) + \Pr(F)\Pr(R \mid F)}\\ & = \frac{\frac{1}{2} \cdot \frac{1}{2}}{\frac{1}{2} \cdot \frac{1}{2} + \frac{1}{2} \cdot 1}\\ & = \frac{\frac{1}{4}}{\frac{1}{4} + \frac{1}{2}}\\ & = \frac{\frac{1}{4}}{\frac{3}{4}}\\ & = \frac{1}{3} \end{align*}

N. F. Taussig
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  • If I may ask with continuing question. if we return the red ball to the box. Now we back again we two balls in box. Red and the other we don't know. We draw from the box and we get again a red ball for second time. Same question, What's the probability the other ball is white? I started with Pr[B|R] where B is the event of drawing red ball a second time. But it's difficult to figure out Pr[B∩R] – Mustafa Shama Nov 22 '22 at 14:28
  • If I understand you correctly, you want to find the probability that the original ball was white given that a red ball was drawn twice when the box contained the red ball and the unknown original ball. I would set up that problem as follows, using the notation I used above: $$\frac{\Pr(E)\Pr(R \mid E)\Pr(R \mid E \cap R)}{\Pr(E)\Pr(R \mid E)\Pr(R \mid E \cap R) + \Pr(F)\Pr(R \mid F)\Pr(R \mid R \cap F)}$$ Observe that since the same two balls are in the box, $\Pr(R \mid E \cap R) = \Pr(R \mid E)$ and $\Pr(R \mid F \cap R) = \Pr(R \mid F)$. – N. F. Taussig Nov 22 '22 at 15:16
  • If you do that calculation, you should find that the probability that the original ball was white is smaller than $1/3$ since you have an additional piece of information that suggests that the ball originally placed in the box was red. – N. F. Taussig Nov 22 '22 at 15:18
  • The events E,F and R are depended right? – Mustafa Shama Nov 22 '22 at 22:05
  • For example, if we put draw the ball and the put it back again in box n times. And all the n times we get red ball each time. What's the probability the original ball is white? For sure the probability will be much smaller – Mustafa Shama Nov 22 '22 at 22:46
  • Yes, those events are dependent. You are also correct that as the number of times you randomly select a ball from the box without obtaining a white ball increases, the conditional probability that a white ball was originally placed in the box decreases. That said, it is possible to draw a red ball ten times in a row even if there is a white ball in the box. The only way you can be certain that there is no white ball in the box is to remove both of the balls and inspect them. – N. F. Taussig Nov 22 '22 at 23:00
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I look at it as a sequence of independent events probability.

  1. put in ball 1 as red (50%) or white (50%).
  2. put in ball 2 as red.
  3. pull out the first ball (50%), or pull out the second ball (50%).

This gives us 4 possible outcomes.

  • (ball1=red, pulled=1) color of pulled ball = red
  • (ball1=red, pulled=2) color of pulled ball = red
  • (ball1=white, pulled=1) color of pulled ball = white
  • (ball1=white, pulled=2) color of pulled ball = red

Only three of these are possible, due to observed ball being red. Of these three cases only one has ball1=white.

So our chance is 1/3

Especially Lime
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Michael Anderson
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