Bag A contains 3 white, 7 red balls and bag B contains 3 white, 2 red balls. One bag is selected at

Q: Wait, why do we use $1/2$ for the probability of picking a bag? The problem doesn't explicitly say it's fair.

In probability problems, if the selection process is described as 'at random' without further qualification, we assume a uniform distribution. Since there are two bags, each has an equal chance of $1/2$.

Q: Why can't I just compare the number of white balls? Bag A has 3 and Bag B has 3. Shouldn't it be 50-50?

That is a classic trap! You are ignoring the total number of balls in each bag. Bag A has 10 balls total, while Bag B has only 5. The 'density' of white balls is higher in Bag B. Bayes' Theorem accounts for this by weighting the probabilities correctly.

Q: Is Bayes' Theorem just a fancy way of doing conditional probability?

Essentially, yes! It is the bridge between the probability of an effect given a cause and the probability of a cause given an effect. It is the mathematical backbone of machine learning and medical diagnosis.

Question

Bag A contains 3 white, 7 red balls and bag B contains 3 white, 2 red balls. One bag is selected at random and a ball is drawn from it. The probability of drawing the ball from the bag A, if the ball drawn in white, is :

Options:

(A) $\frac{1}{4}$

(B) $\frac{1}{9}$

(C) $\frac{1}{3}$

(D) $\frac{3}{10}$

Sigma Percentile · Accepted Answer

### The Mystery of the Two Bags: A Journey into Bayes' Theorem

Imagine you are standing in a quiet room with two bags in front of you. Bag A, on your left, contains 3 white balls and 7 red balls, for a total of 10. Bag B, on your right, holds 3 white balls and 2 red balls, for a total of 5.

You are asked to pick one bag at random and draw a ball. Upon drawing, you observe a white ball. The question is: what is the probability that this ball came from Bag A?

This is a classic exercise in reverse-thinking, a core skill for any aspiring physicist or engineer. We are looking for the probability of a cause (Bag A) given an observed effect (a white ball).

### Defining Our Universe

To solve this, we must first map out our universe. Let $E_1$ be the event of choosing Bag A, and $E_2$ be the event of choosing Bag B.

Since the selection is random, we assign equal weight to both:
$$P(E_1) = \frac{1}{2}, \quad P(E_2) = \frac{1}{2}$$

Now, let $E$ be the event of drawing a white ball. This is our 'evidence'. We need to calculate the conditional probability $P(E_1|E)$, which represents the probability of $E_1$ given $E$.

### The Likelihoods

Before we can use Bayes' Theorem, we need the 'likelihoods'—the probability of the evidence given each hypothesis. If we were holding Bag A, the chance of drawing a white ball is:
$$P(E|E_1) = \frac{3}{10}$$

If we were holding Bag B, the chance is:
$$P(E|E_2) = \frac{3}{5}$$

Notice how Bag B has a higher concentration of white balls. This intuition will be crucial when we see the final result.

### The Bayes' Theorem Engine

Bayes' Theorem is the mathematical bridge that allows us to reverse the conditional probability. It is defined as:

$$P(E_1|E) = \frac{P(E_1)P(E|E_1)}{P(E_1)P(E|E_1) + P(E_2)P(E|E_2)}$$

Think of the numerator as the 'favorable path'—the probability of choosing Bag A AND getting a white ball. The denominator is the 'total probability' of getting a white ball, which includes the path through Bag A and the path through Bag B.

### The Calculation

Let us plug in our numbers. The numerator is:
$$\frac{1}{2} 	imes \frac{3}{10} = \frac{3}{20}$$

The denominator is the sum of the two paths:
$$\left(\frac{1}{2} 	imes \frac{3}{10}ight) + \left(\frac{1}{2} 	imes \frac{3}{5}ight) = \frac{3}{20} + \frac{3}{10}$$

To add these, we find a common denominator:
$$\frac{3}{20} + \frac{6}{20} = \frac{9}{20}$$

Finally, we calculate the conditional probability:
$$P(E_1|E) = \frac{3/20}{9/20} = \frac{3}{9} = \frac{1}{3}$$

### The Conclusion

Despite Bag A having the same number of white balls as Bag B, the probability that the ball came from Bag A is only **$1/3$**.

This occurs because Bag B is 'denser' with white balls, making it a more likely source for the white ball we observed. This is the beauty of probability—it forces us to look past our initial assumptions and calculate the truth based on the evidence.

Animated Solution for Mathematics - Probability: Bag A contains 3 white, 7 red balls and bag B contains 3 white, 2 red balls. One bag is selected at random and a ball is drawn from it. The probability of drawing the ball from the bag A, if the ball drawn in white, is :

Select Answer:

Visualized Solution

The Physical Setup

The Observation

Defining the Events

Prior Probabilities

Conditional Probabilities

Applying Bayes' Theorem

Substituting the Values

Simplifying the Products

Adding the Denominator

Final Probability

The Sigma Insight: Bayes' Theorem

The Mystery of the Two Bags

Defining Our Universe

The Likelihoods

The Bayes' Theorem Engine

The Calculation

The Conclusion

Similar Questions

Bag $B_{1}$ contains 6 white and 4 blue balls, Bag $B_{2}$ contains 4 white and 6 blue balls, and Bag $B_{3}$ contains 5 white and 5 blue balls. One of the bags is selected at random and a ball is drawn from it. If the ball is white, then the probability, that the ball is drawn from Bag $B_{2}$ is:

Bag I contains 3 red, 4 black and 3 white balls and Bag II contains 2 red, 5 black and 2 white balls. One ball is transferred from Bag I to Bag II and then a ball is draw from Bag II. The ball so drawn is found to be black in colour. Then the probability, that the transferred ball is red, is:

Bag A contains 2 white, 1 black and 3 red balls and bag B contains 3 black, 2 red and $n$ white balls. One bag is chosen at random and 2 balls drawn from it at random, are found to be 1 red and 1 black. If the probability that both balls come from Bag A is $\frac{6}{11}$ , then $n$ is equal to ____.

A bag contains 8 balls, whose colours are either white or black. 4 balls are drawn at random without replacement and it was found that 2 balls are white and other 2 balls are black. The probability that the bag contains equal number of white and black balls is:

A bag contains 10 balls out of which $k$ are red and $(10 - k)$ are black, where $0 \leq k \leq 10$ . If three balls are drawn at random without replacement and all of them are found to be black, then the probability that the bag contains 1 red and 9 black balls is:

Comprehension Passage

If 1 ball is drawn from each of the boxes $B_{1}, B_{2}$ and $B_{3}$ , the probability that all 3 drawn balls are of the same colour is

If 2 balls are drawn (without replacement) from a randomly selected box and one of the balls is white and the other ball is red, the probability that these 2 balls are drawn from box $B_{2}$ is

A bag contains 6 balls. Two balls are drawn from it at random and both are found to be black. The probability that the bag contains at least 5 black balls is