There are two boxes I and II. Box I contains 3 red and 6 black balls. Box II contains 5 red and 5 black balls. One of the two boxes, box I and box II, is selected at random and a ball is drawn at random. The ball drawn is found to be red. Find the probability that this red ball c

Question

There are two boxes I and II. Box I contains 3 red and 6 black balls. Box II contains 5 red and 5 black balls. One of the two boxes, box I and box II, is selected at random and a ball is drawn at random. The ball drawn is found to be red. Find the probability that this red ball comes out from box II.

Accepted Answer

> [!TLDR]
> This is a classic Bayes' theorem problem. We are given the prior probability of selecting each box (1/2 each) and the conditional probability of drawing a red ball from each box. The probability that the red ball came from Box II is $\frac{5}{8}$.

The key idea here is **conditional probability** — specifically, we are asked to "reverse" the direction of the given information. We know the probability of drawing a red ball *given* a particular box. But we need the probability that the box was Box II *given* that the ball drawn is red. This reversal is exactly what Bayes' theorem handles.

Think of it this way: before seeing the ball, each box is equally likely. But once we see a red ball, that evidence updates our belief. Box II has a higher proportion of red balls (5 out of 10) than Box I (3 out of 9), so a red ball is more likely to have come from Box II. Bayes' theorem lets us quantify exactly how much more likely.

Let's work through it step by step.

1. **Define the events clearly.**  
   Let $B_1$ be the event that Box I is selected, and $B_2$ the event that Box II is selected.  
   Let $R$ be the event that a red ball is drawn.

2. **Write down the prior probabilities.**  
   Since one box is selected at random from the two:  
   $$P(B_1) = \frac{1}{2}, \quad P(B_2) = \frac{1}{2}$$

3. **Write down the conditional probabilities (likelihoods).**  
   Box I has 3 red out of 9 total balls, so:  
   $$P(R \mid B_1) = \frac{3}{9} = \frac{1}{3}$$  
   Box II has 5 red out of 10 total balls, so:  
   $$P(R \mid B_2) = \frac{5}{10} = \frac{1}{2}$$

4. **Find the total probability of drawing a red ball.**  
   By the law of total probability:  
   $$P(R) = P(R \mid B_1) P(B_1) + P(R \mid B_2) P(B_2)$$  
   Substituting:  
   $$P(R) = \left(\frac{1}{3}\right)\left(\frac{1}{2}\right) + \left(\frac{1}{2}\right)\left(\frac{1}{2}\right) = \frac{1}{6} + \frac{1}{4}$$  
   To add, find a common denominator (12):  
   $$P(R) = \frac{2}{12} + \frac{3}{12} = \frac{5}

Concept understanding — Conditional Probability

Conditional Probability

The Idea: Shrink the Sample Space

The Definition

Two Cautions

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