# Probability question (simple)

Consider two boxes:

• The first box contains $4$ red balls and $2$ black balls.
• The second box contains $5$ red balls and $3$ white balls.

Next we throw a dice.

• If the result is an even number, we draw from the first box
• If the result is an odd number, draw from the second box

If a draw ball is red, what is the probability that it was drawn from 1st box?

• I've tried to reformulate the question as it was hard to read the way you wrote it down. Could you please confirm that this is what you meant to ask... – gebruiker Jun 11 '14 at 17:25
• yes, is correct – techoo Jun 11 '14 at 17:29

This is a standard conditional probability problem. One could solve it informally, by drawing a tree diagram. But instead, we will set up and use some formal machinery.

Let $R$ be the event that the ball is red. Let $F$ be the event it was drawn from the first box. We want the conditional probability $\Pr(F|R)$, the probability that the ball is from the first box given that it is red.

By the defining formula for conditional probability, we have $$\Pr(F|R)=\frac{\Pr(F\cap R)}{\Pr(R)}.$$ We compute the two probabilities on the right.

The ball can be red in two ways: (i) we drew from the first box, and the result was red or (ii) we drew from the second box, and the ball was red.

The probability we drew from the first box is $\displaystyle\frac{3}{6}$. Given we drew from the first box, the probability of red is $\displaystyle\frac{4}{6}$. so the probability of (i) is $\displaystyle\frac{3}{6}\cdot\frac{4}{6}$.

Similarly, the probability of (ii) is $\displaystyle\frac{3}{6}\cdot\frac{5}{8}$.

For $\Pr(R)$, add our two numbers.

Finally, note that $\Pr(F\cap R)$ is just the probability of (i). Now you have all the ingredients to find the answer.

Remark: The following is imprecise, but may supply some intuition. suppose we do the experiment $1200$ times. Then roughly $600$ times we will pick from the first box, and roughly $600$ times we will pick from the second.

Out of the about $600$ times we pick from the first, we will get about $400$ reds. Out of the about $600$ we pick from the second, we will get about $375$ reds, for a total of $775$. In $400$ of the cases, the red came from the first box. So the probability we picked from the first box given we got red should be about $\frac{400}{775}$.

2 steps:

1. $P(1st|red)=\frac{P(\text{red from 1st})}{P(red)}$
2. $P(red)=P(\text{red from 1st})P(1st)+P(\text{red from 2nd})P(2nd)$