Suppose that Box-I contains 8 red, 3 blue and 5 green balls, Box-II contains 24 red, 9 blue and 15 g

Q: Wait, why is $P(W \cap G) = P(W)$? Does this mean $G$ is always true if $W$ is true?

Exactly! Think about the condition for $W$ (drawing a white ball). A white ball is only in Box-IV. To draw from Box-IV, the first ball must be Green. If the first ball is Green, the condition for $G$ (at least one green ball) is already satisfied. Thus, $W$ is a subset of $G$, and the intersection is just $W$.

Q: Why don't we care about the second ball in Case 3 (where the first ball is Green)?

Great observation! The event $G$ is 'at least one green ball'. If the first ball is Green, you have already satisfied the condition. It doesn't matter what the second ball is; the event $G$ has already occurred. So, we just take the probability of picking a Green ball from Box-I.

Q: Is there a simpler way to calculate $P(G)$ using the complement?

You could try $1 - P(G^c)$, where $G^c$ is the event 'no green balls'. However, because the boxes have different compositions, you would still need to calculate the probability of picking a non-green ball from each box based on the first draw. The direct method used here is actually more straightforward and less prone to calculation errors!

Question

Suppose that Box-I contains 8 red, 3 blue and 5 green balls, Box-II contains 24 red, 9 blue and 15 green balls, Box-III contains 1 blue, 12 green and 3 yellow balls, Box-IV contains 10 green, 16 orange and 6 white balls. A ball is chosen randomly from Box-I; call this ball $b$. If $b$ is red then a ball is chosen randomly from Box-II, if $b$ is blue then a ball is chosen randomly from Box-III, and if $b$ is green then a ball is chosen randomly from Box-IV. The conditional probability of the event 'one of the chosen balls is white' given that the event 'at least one of the chosen balls is green' has happened, is equal to

Options:

(A) $\frac{15}{256}$

(B) $\frac{3}{16}$

(C) $\frac{5}{52}$

(D) $\frac{1}{8}$

Sigma Percentile · Accepted Answer

### The Probability Forest: A Journey Through Dependent Events

Welcome, fellow traveler of the mathematical landscape. Today, we are not just solving a probability problem; we are navigating a forest of branching paths.

When we face a problem involving multiple boxes and sequential draws, the most powerful tool in our arsenal is the tree diagram. It allows us to visualize the dependency: the first choice we make dictates the entire future of our experiment.

### Phase 1: The Setup and the Goal

We start with four boxes, each with a distinct composition. Let's list them out to keep our bearings:

- **Box-I**: $8R, 3B, 5G$ (Total = $16$)
- **Box-II**: $24R, 9B, 15G$ (Total = $48$)
- **Box-III**: $1B, 12G, 3Y$ (Total = $16$)
- **Box-IV**: $10G, 16O, 6W$ (Total = $32$)

The problem asks for the conditional probability $P(W|G)$, where $W$ is the event that one chosen ball is White, and $G$ is the event that at least one chosen ball is Green. By the fundamental definition of conditional probability, we know that:

$$P(W|G) = \frac{P(W \cap G)}{P(G)}$$

Our mission is clear: calculate the numerator and the denominator separately, then bring them together for the final result.

### Phase 2: The Intersection $P(W \cap G)$

Let's analyze the numerator. Where can we find a white ball? Only in Box-IV.

To draw from Box-IV, the first ball drawn from Box-I must be Green. If the first ball is Green, we have already satisfied the condition for event $G$ (at least one green ball).

Therefore, the event $W$ (drawing a white ball) is a subset of $G$. This simplifies our life immensely: $P(W \cap G) = P(W)$.

To calculate $P(W)$, we follow the path: Green from Box-I, then White from Box-IV:

$$P(W) = P(G_1) 	imes P(W_2 | G_1) = \frac{5}{16} 	imes \frac{6}{32} = \frac{15}{256}$$

### Phase 3: The Total Probability $P(G)$

Now, for the denominator. We need the total probability of drawing at least one green ball. This can happen in three mutually exclusive ways, depending on the first draw:

1. **First is Red**: We go to Box-II. We need a Green ball from Box-II.

$$P(R_1 \cap G_2) = P(R_1) 	imes P(G_2 | R_1) = \frac{8}{16} 	imes \frac{15}{48} = \frac{5}{32}$$

2. **First is Blue**: We go to Box-III. We need a Green ball from Box-III.

$$P(B_1 \cap G_3) = P(B_1) 	imes P(G_3 | B_1) = \frac{3}{16} 	imes \frac{12}{16} = \frac{9}{64}$$

3. **First is Green**: We have already satisfied the condition! We don't need to worry about the second ball.

$$P(G_1) = \frac{5}{16}$$

Summing these up to find $P(G)$:

$$P(G) = \frac{5}{32} + \frac{9}{64} + \frac{5}{16} = \frac{10 + 9 + 20}{64} = \frac{39}{64}$$

### Phase 4: The Final Synthesis

We have our numerator, $\frac{15}{256}$, and our denominator, $\frac{39}{64}$. Now, we perform the final division:

$$P(W|G) = \frac{\frac{15}{256}}{\frac{39}{64}} = \frac{15}{256} 	imes \frac{64}{39}$$

Simplifying this, we see that $64$ goes into $256$ four times, and $15$ and $39$ share a factor of $3$:

$$P(W|G) = \frac{15}{4 	imes 39} = \frac{5}{4 	imes 13} = \mathbf{\frac{5}{52}}$$

And there we have it! By breaking the problem into logical, manageable paths, we navigated the forest and arrived at the solution. Remember, in probability, the complexity is often just a collection of simple steps waiting to be organized.

Select Answer:

Visualized Solution

The Setup

Defining the Goal

Analyzing Event $W$

The Intersection $W \cap G$

Calculating $P (W)$

Analyzing Event $G$

Case 1: First is Red

Case 2: First is Blue

Case 3: First is Green

Total Probability $P (G)$

Final Substitution

Final Answer

The Sigma Insight: Conditional Probability

The Probability Forest

Phase 1

Phase 2

Phase 3

Phase 4

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Select Answer:

Visualized Solution

The Setup

Defining the Goal

Analyzing Event .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } W

The Intersection .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } W∩G

Calculating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P(W)

Analyzing Event .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } G