There are three bags B₁, B₂ and B₃. The bag B₁ contains 5 red and 5 green balls, B₂ contains 3 red a

Q: Wait, why is $P(G|B_3)$ just $\frac{3}{8}$? Shouldn't I multiply by the probability of picking Bag 3?

Great question! That is the classic trap. $P(G|B_3)$ is a *conditional* probability—it assumes Bag 3 is *already* in your hand. You do not multiply by $P(B_3)$ until you are calculating the *joint* probability, $P(G \cap B_3)$.

Q: Why does Bayes' Theorem feel so counter-intuitive? It is like working backward in time.

Exactly! That is why it is powerful. You are essentially 'filtering' the sample space. You are ignoring all scenarios where a red ball was picked and looking only at the 'green ball' universe to see what fraction of that universe came from Bag 3.

Q: Why do we use the denominator $80$ for everything? Is there a trick?

It is not a trick, but a strategic choice! When dealing with fractions like $\frac{3}{10}$, $\frac{5}{8}$, and $\frac{3}{8}$, finding a common denominator (which is $80$ here) makes the addition of probabilities much cleaner and less prone to arithmetic errors.

Question

There are three bags $B_1, B_2$ and $B_3$. The bag $B_1$ contains 5 red and 5 green balls, $B_2$ contains 3 red and 5 green balls, and $B_3$ contains 5 red and 3 green balls. Bags $B_1, B_2$ and $B_3$ have probabilities $\frac{3}{10}, \frac{3}{10}$ and $\frac{4}{10}$ respectively of being chosen. A bag is selected at random and a ball is chosen at random from the bag. Then which of the following options is/are correct?

Options:

(A) Probability that the selected bag is $B_3$ and the chosen ball is green equals $\frac{3}{10}$

(B) Probability that the chosen ball is green equals $\frac{39}{80}$

(C) Probability that the chosen ball is green, given that the selected bag is $B_3$, equals $\frac{3}{8}$

(D) Probability that the selected bag is $B_3$, given that the chosen balls is green, equals $\frac{5}{13}$

Sigma Percentile · Accepted Answer

### The Probability Odyssey: Unlocking the Mystery of the Three Bags

Welcome, future engineer! Today, we are diving into the elegant world of probability. This problem is not just about crunching numbers; it is about understanding the architecture of chance.

Imagine you are standing in a room with three distinct bags, $B_1$, $B_2$, and $B_3$. Each bag holds a different secret—a different ratio of red and green balls. Your task is to navigate this uncertainty using the power of logic.

### The Anatomy of the Experiment

Before we calculate anything, let us visualize the process. We have three bags, and each has a specific probability of being chosen:
$P(B_1) = \frac{3}{10}$, $P(B_2) = \frac{3}{10}$, and $P(B_3) = \frac{4}{10}$.

Notice that Bag 3 is slightly more likely to be picked. This is our starting point.

Now, consider the contents:
*   Bag 1 has $5$ red and $5$ green balls.
*   Bag 2 has $3$ red and $5$ green balls.
*   Bag 3 has $5$ red and $3$ green balls.

We are specifically hunting for a green ball, which we will call event $G$.

### The Conditional Reality

Let us tackle Option 3 first. It asks for $P(G|B_3)$, the probability of drawing a green ball *given* that Bag 3 is already in your hand.

This is a conditional probability. We do not care about the probability of picking the bag here; we assume the bag is already picked.

Looking inside Bag 3, we see $3$ green balls out of a total of $8$ balls ($5$ red + $3$ green). Thus:
$$P(G|B_3) = \frac{3}{8}$$

This confirms that Option 3 is correct. It is a simple, direct observation of the bag's contents.

### The Total Probability

Now, things get interesting. Option 2 asks for the total probability of drawing a green ball, $P(G)$. Since a green ball can emerge from any of the three bags, we must account for all possible paths using the Law of Total Probability.

We sum the probabilities of drawing a green ball from each bag, weighted by the probability of choosing that bag:
$$P(G) = P(B_1)P(G|B_1) + P(B_2)P(G|B_2) + P(B_3)P(G|B_3)$$

Let us plug in our values:
$$P(G) = \left(\frac{3}{10} 	imes \frac{5}{10}ight) + \left(\frac{3}{10} 	imes \frac{5}{8}ight) + \left(\frac{4}{10} 	imes \frac{3}{8}ight)$$

Calculating these terms, we get:
$$P(G) = \frac{15}{100} + \frac{15}{80} + \frac{12}{80}$$

Simplifying $\frac{15}{100}$ to $\frac{3}{20}$ and converting it to $\frac{12}{80}$, we have:
$$\frac{12}{80} + \frac{15}{80} + \frac{12}{80} = \frac{39}{80}$$

This matches Option 2 perfectly. We have successfully navigated the entire probability space!

### The Reverse Logic (Bayes' Theorem)

Finally, let us look at Option 4. It asks for $P(B_3|G)$, the probability that Bag 3 was selected *given* that we have already drawn a green ball. This is the heart of Bayes' Theorem—the 'reverse' probability.

The formula is:
$$P(B_3|G) = \frac{P(B_3 \cap G)}{P(G)}$$

We already know $P(G) = \frac{39}{80}$. The numerator, $P(B_3 \cap G)$, is simply the probability of picking Bag 3 AND getting a green ball:
$$P(B_3 \cap G) = P(B_3) 	imes P(G|B_3) = \frac{4}{10} 	imes \frac{3}{8} = \frac{12}{80}$$

Dividing these, we get:
$$\frac{12/80}{39/80} = \frac{12}{39} = \frac{4}{13}$$

Option 4 claims the answer is $\frac{5}{13}$, so it is **incorrect**. Through this journey, we have seen how conditional probability, total probability, and Bayes' theorem weave together to solve complex problems. Keep practicing, and these concepts will become second nature!

Select Answer:

Visualized Solution

Problem Setup: The Bags

Problem Setup: The Balls

Checking Option 3: $P (G ∣ B_{3})$

Checking Option 1: $P (B_{3} \cap G)$

Evaluating Option 1

Checking Option 2: Total Probability

Substituting Values for Total Probability

Multiplying the Terms

Adding the Probabilities

Checking Option 4: $P (B_{3} ∣ G)$

Substituting Values into Bayes' Theorem

Evaluating Option 4

Final Conclusion

The Sigma Insight: Bayes' Theorem

The Probability Odyssey

The Anatomy of the Experiment

The Conditional Reality

The Total Probability

The Reverse Logic (Bayes' Theorem)

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 :

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:

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

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)

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 ____.

Three urns A, B and C contain 7 red, 5 black; 5 red, 7 black and 6 red, 6 black balls, respectively. One of the urn is selected at random and a ball is drawn from it. If the ball drawn is black, then the probability that it is drawn from urn A is :

Select Answer:

Visualized Solution

Problem Setup: The Bags

Problem Setup: The Balls

Checking Option 3: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P(G∣B3​)

Checking Option 1: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P(B3​∩G)

Evaluating Option 1

Checking Option 2: Total Probability

Substituting Values for Total Probability

Multiplying the Terms

Adding the Probabilities

Checking Option 4: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P(B3​∣G)

Substituting Values into Bayes' Theorem

Evaluating Option 4

Final Conclusion

The Sigma Insight: Bayes' Theorem

The Probability Odyssey

The Anatomy of the Experiment

The Conditional Reality

The Total Probability

The Reverse Logic (Bayes' Theorem)

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 :

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:

Comprehension Passage

.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

Comprehension Passage

Three urns A, B and C contain 7 red, 5 black; 5 red, 7 black and 6 red, 6 black balls, respectively. One of the urn is selected at random and a ball is drawn from it. If the ball drawn is black, then the probability that it is drawn from urn A is :

Checking Option 3: $P (G ∣ B_{3})$

Checking Option 1: $P (B_{3} \cap G)$

Checking Option 4: $P (B_{3} ∣ G)$

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