# What is the probability of the given event?

Consider 2 Boxes. One has 99 white balls and 1 red ball. Another box has 99 red balls and 1 white ball. Now one has to chose a box at random and take out a red ball. It is given that the person has taken out the red ball. Now what is the probability that the ball has been chosen from box 1 which has 99 white balls and 1 red ball.

What I think is that no matter what box one chose, the final result is a red ball. And since it is given that the ball indeed was red, therfore the probability that red ball is from box 1 is 0.5 since in the beginning one didn't know which box has how many balls. Moreover once the box was chosen, the probability that the red ball is going to come was 1 since it is given the ball taken out was red.

• Are you familiar with Bayes' Theorem? Commented Apr 13, 2022 at 15:48
• This is pretty much an application of conditional probabilities. You (should) know that $P(A \mid B) P(B) = P(A \cap B) = P(B \mid A) P(A),$ the equality between the two extremes is sometimes called "Bayes' theorem" (a misnomer since Bayes himself never formulated such equation). Setting $A$ to be "Box one was chosen" and $B$ "A red ball was drawn" you can substitute and find the desired probability. Note that $P(A) = 1/2$ by assumtion and $P(B) = 1/2$ since there is a total of 100 red balls out of 200 balls. Commented Apr 13, 2022 at 16:18
• Another intuitive way of handling this is the following: there are as many red as white balls, therefore, the odds are 1-to-1 for either colour. So, the chances of getting a box or the other is proportional to the number of balls inside them, which is therefore 1/100. Commented Apr 13, 2022 at 16:30
• Please take a look at my answer here (be sure to examine the tree diagram) to "A bag I contain 4 white and 6 black balls while another Bag II contains 4 white and 3 black balls. One ball is drawn at random from one of the bags, and it is found to be black. Find the probability that it was drawn from Bag I." Commented Apr 13, 2022 at 16:45

This is a simple question if you are familiar with the concept of conditional probability. Given two events $$A$$ and $$B$$, we will write the probability that the event $$B$$ is true as $$p_{A|B}$$. For instance in your case $$A$$ is the event "the ball is drawn from the box 1" and $$B$$ is "a red ball is drawn". You need to find the value of $$p_{A|B}$$.

First, so far we know $$p_A$$, that is the probability of choosing the box 1, and $$p_{B|A}$$, that is once we have chosen the box 1 what is the probability of extracting a red ball. It is quite intuitive to see that extracting a red ball from the box one (that is having both the events $$A$$ and $$B$$) can occur if and only if first I choose the box 1 (which will occur with probability $$p_A$$) and then I draw the red ball (which, given that we have chosen the box 1 will occur with probability $$p_{B|A}$$). From this it should make sense that $$p_{A, B} = p_A p_{B|A}$$, that is the probability $$p_{A, B}$$ of extracting a red ball from the box one is exactly the product of $$p_A$$ (first I choose the box 1) and $$p_{B|A}$$ (now I extract the red ball from the box 1).

From all we have done above we got that $$p_{A, B} = p_A p_{B|A}\,.$$ Note that the power of the above formula is its abstraction. It does not really matter what the events $$A$$ and $$B$$ means, the formula will be true. In particular, we can swap $$A$$ and $$B$$ and write $$p_{A, B} = p_B p_{A|B}\,.$$

From here you get that $$p_{A|B}$$, which is what you are actually interested in, is given by $$p_{A|B} = \frac{p_{A,B}}{p_B} = \frac{p_A p_{B|A}}{p_B}\,.$$

Now you know everything but $$p_B$$, that is the probability of extracting a red ball. How to evaluate it? Well it should make sense to say that it corresponds to the probability of choosing the box 1 ($$p_A$$) times the probability of extracting a red ball from there ($$p_{B|A}$$), plus the probability of choosing the box 2 ($$p_{A^c} = 1-p_A$$) times the probability of extracting a red ball from there ($$p_{B|A^c}$$), where $$A^c$$ is the event "choxsing the box 2". This means $$p_B = p_A p_{B|A} + p_{A^c}p_{B|A^c}\,.$$

In your case $$p_A = p_{A^c} = 1/2$$, $$p_{B|A} = 1/100$$ and $$p_{B|A^c} = 99/100$$, so $$p_B = 1/2$$. We can thus find $$p_{A|B} = \frac{1/2\times 1/100}{1/2} = 1/100\,.$$

Some comments. So what was it wrong about your intuition? When would the result be $$1/2$$. Well your answer would be correct in the following scenario. Somebody is extracting balls from one of the two boxes. She tells you nothing until she finally extract a red ball. At that moment she tells you that she has a red ball and you have to guess the box from which she drawn it. Then indeed you have no way to guess and the probability of both the box 1 and 2 will be $$1/2$$. The fundamental difference here is that in your problem there was one single draw and it was successful, here she is extracting until a red ball appears and then telling you about the red ball.

A situation similar to yours that maybe might sound more intuitive can be the following. Let's say that there are two classes with the same number of people in a school, in one class people study Italian, in the other they don't. Yet there is a very bad student in the Italian studying class who cannot say a single word of Italian, while in the other class one student comes from an Italian family and thus can speak Italian. Imagine that you meet one student outside of this school, and he can speak Italian. From which classes would you guess he comes?