Bag A contains 2 white, 1 black and 3 red balls and bag B contains 3 black, 2 red and n white balls.

Q: Wait, why do we use Bayes' Theorem here instead of just simple probability?

Great question! Simple probability works when we know the cause and want to predict the effect. Here, we know the effect (the balls drawn are 1 red and 1 black) and we are trying to determine the cause (which bag they came from). This 'reverse' logic is exactly what Bayes' Theorem is designed to handle.

Q: How do I handle the $n$ in the denominator when calculating the combinations for Bag B?

Don't let the variable intimidate you! The total number of balls in Bag B is $n+5$. When you draw 2 balls, the total ways to do this is $^{n+5}C_2$, which expands to $\frac{(n+5)(n+4)}{2}$. When you put this in the denominator of a fraction, the $2$ will flip to the numerator, making your algebra much cleaner.

Q: Why do we reject the negative root when solving the quadratic equation?

In physics and probability problems, variables often represent physical quantities—in this case, the number of balls in a bag. Since you cannot have a negative number of balls, any negative solution for $n$ is physically impossible and must be discarded. Always check your final answer against the physical constraints of the problem!

Question

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 ____.

Options:

(A) 13

(B) 6

(C) 4

(D) 3

Sigma Percentile · Accepted Answer

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

Imagine you are standing in a room with two bags, Bag A and Bag B. You don't know what's inside Bag B, but you know Bag A has 2 white, 1 black, and 3 red balls.

You reach in, pick a bag at random, and draw two balls. To your surprise, you pull out exactly 1 red and 1 black ball.

Now, the question is: what is the probability that these balls came from Bag A? And more importantly, if we know that probability is $\frac{6}{11}$, can we figure out how many white balls were in Bag B?

This is the essence of **Bayes' Theorem**—a powerful tool that allows us to update our beliefs based on new evidence.

### Phase 1: Setting the Stage

First, let's define our universe. We have two events: $E_1$ (choosing Bag A) and $E_2$ (choosing Bag B).

Since the bag is chosen at random, the probability of picking either is equal: 
$$P(E_1) = P(E_2) = \frac{1}{2}$$

Now, let's look at the contents. Bag A has a total of $2+1+3 = 6$ balls. Bag B has 3 black, 2 red, and $n$ white balls, for a total of $n+5$ balls.

We define Event $A$ as the outcome: drawing 1 red and 1 black ball.

### Phase 2: Calculating the Conditional Probabilities

To use Bayes' Theorem, we need to know the probability of drawing our specific balls from each bag individually.

For Bag A, the probability $P(A|E_1)$ is the number of ways to pick 1 red and 1 black divided by the total ways to pick 2 balls from 6: 
$$P(A|E_1) = \frac{{}^3C_1 	imes {}^1C_1}{{}^6C_2} = \frac{3 	imes 1}{15} = \frac{1}{5}$$

For Bag B, the probability $P(A|E_2)$ is slightly more complex because of the variable $n$: 
$$P(A|E_2) = \frac{{}^2C_1 	imes {}^3C_1}{{}^{n+5}C_2} = \frac{2 	imes 3}{\frac{(n+5)(n+4)}{2}} = \frac{12}{(n+5)(n+4)}$$

### Phase 3: The Bayes' Bridge

Bayes' Theorem tells us that the probability of having chosen Bag A, given that we observed Event $A$, is: 
$$P(E_1|A) = \frac{P(E_1)P(A|E_1)}{P(E_1)P(A|E_1) + P(E_2)P(A|E_2)}$$

We know $P(E_1|A) = \frac{6}{11}$. Substituting our values, we get: 
$$\frac{\frac{1}{2} \cdot \frac{1}{5}}{\frac{1}{2} \cdot \frac{1}{5} + \frac{1}{2} \cdot \frac{12}{(n+5)(n+4)}} = \frac{6}{11}$$

### Phase 4: The Algebraic Dance

This looks intimidating, but notice the beauty of the symmetry! The $\frac{1}{2}$ term appears in every single part of the expression. We can factor it out and cancel it entirely: 
$$\frac{\frac{1}{5}}{\frac{1}{5} + \frac{12}{(n+5)(n+4)}} = \frac{6}{11}$$

To simplify further, multiply the numerator and denominator by 5: 
$$\frac{1}{1 + \frac{60}{(n+5)(n+4)}} = \frac{6}{11}$$

Now, take the reciprocal of both sides: 
$$1 + \frac{60}{(n+5)(n+4)} = \frac{11}{6}$$

Subtracting 1 from both sides gives: 
$$\frac{60}{(n+5)(n+4)} = \frac{5}{6}$$

### Phase 5: The Final Resolution

Cross-multiplying, we find: 
$$(n+5)(n+4) = \frac{60 	imes 6}{5} = 72$$

Expanding the left side: 
$$n^2 + 9n + 20 = 72 \Rightarrow n^2 + 9n - 52 = 0$$

Factoring this quadratic equation, we look for two numbers that multiply to $-52$ and add to $9$. Those numbers are $13$ and $-4$: 
$$(n+13)(n-4) = 0$$

This gives us two possible values: $n = -13$ or $n = 4$. Since $n$ represents the number of balls, it must be positive.

Thus, we reject $-13$ and conclude that **$n = 4$**. We have successfully solved the mystery of the bags!

Select Answer:

Visualized Solution

Visualizing the Bags

Defining Bag Selection Events

Defining the Outcome Event $A$

Probability from Bag A: $P (A ∣ E_{1})$

Probability from Bag B: $P (A ∣ E_{2})$

Applying Bayes' Theorem

Substituting the Values

Canceling Common Terms

Simplifying the Equation

Taking the Reciprocal

Solving for $n$

Forming the Quadratic Equation

Factoring the Quadratic

Conclusion & Takeaway

The Sigma Insight: Bayes' Theorem

The Mystery of the Two Bags

Phase 1

Phase 2

Phase 3

Phase 4

Phase 5

Similar Questions

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 :

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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:

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:

Comprehension Passage

A ball is drawn at random from box I and transferred to box II. If the probability of drawing a red ball from box I, after this transfer, is 1/3, then the correct option(s) with the possible values of $n_{1}$ and $n_{2}$ is(are)

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

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

Select Answer:

Visualized Solution

Visualizing the Bags

Defining Bag Selection Events

Defining the Outcome Event .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A

Probability from Bag A: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P(A∣E1​)

Probability from Bag B: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P(A∣E2​)

Applying Bayes' Theorem

Substituting the Values

Canceling Common Terms

Simplifying the Equation

Taking the Reciprocal

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } n

Forming the Quadratic Equation

Factoring the Quadratic

Conclusion & Takeaway

The Sigma Insight: Bayes' Theorem

The Mystery of the Two Bags

Phase 1

Phase 2

Phase 3

Phase 4

Phase 5

Similar Questions

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Comprehension Passage

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

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.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If 1 ball is drawn from each of the boxes B1​,B2​ and B3​, the probability that all 3 drawn balls are of the same colour is

Defining the Outcome Event $A$

Probability from Bag A: $P (A ∣ E_{1})$

Probability from Bag B: $P (A ∣ E_{2})$

Solving for $n$

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