Understanding the Probability Problem

The question asks for a conditional probability. We are given information about drawing a ball (it's white) and we need to find the probability that it came from a specific bag (Bag-1), based on an initial die roll that determined which bag to draw from.

Let's break down the problem into events and their probabilities.

Defining Events and Probabilities

First, consider the die roll:

• Event D3: The die shows a number divisible by 3. The numbers divisible by 3 on a standard six-sided die are 3 and 6.
• Event ND3: The die shows a number not divisible by 3. The numbers not divisible by 3 are 1, 2, 4, and 5.

The probabilities of these events are:

• Probability of D3, \( P(D3) = \frac{2}{6} = \frac{1}{3} \)
• Probability of ND3, \( P(ND3) = \frac{4}{6} = \frac{2}{3} \)

The problem states that if the die shows a number divisible by 3, a ball is drawn from Bag-1. If it shows a number not divisible by 3, a ball is drawn from Bag-2.

• Event B1: A ball is drawn from Bag-1. This happens if D3 occurs. So, \( P(B1) = P(D3) = \frac{1}{3} \).
• Event B2: A ball is drawn from Bag-2. This happens if ND3 occurs. So, \( P(B2) = P(ND3) = \frac{2}{3} \).

Next, consider the contents of the bags and the probability of drawing a white ball from each:

• Bag-1 contains 4 white and 6 black balls. Total balls = 10.
• Bag-2 contains 5 white and 5 black balls. Total balls = 10.

Let W be the event that the ball drawn is white (not black).

• Probability of drawing a white ball given it was drawn from Bag-1, \( P(W|B1) = \frac{\text{Number of white balls in Bag-1}}{\text{Total balls in Bag-1}} = \frac{4}{10} = \frac{2}{5} \).
• Probability of drawing a white ball given it was drawn from Bag-2, \( P(W|B2) = \frac{\text{Number of white balls in Bag-2}}{\text{Total balls in Bag-2}} = \frac{5}{10} = \frac{1}{2} \).

Calculating the Required Conditional Probability

We are given that the ball drawn is not black (i.e., it is white, event W). We need to find the probability that it was not drawn from Bag-2, which means it was drawn from Bag-1 (event B1).

So, we need to calculate \( P(B1 | W) \), the probability that the ball was drawn from Bag-1 given that it is white.

We can use Bayes' Theorem for this:

\( P(B1 | W) = \frac{P(W | B1) \cdot P(B1)}{P(W)} \)

To use this formula, we first need to calculate \( P(W) \), the total probability of drawing a white ball. We can use the Law of Total Probability:

\( P(W) = P(W | B1) \cdot P(B1) + P(W | B2) \cdot P(B2) \)

Substitute the probabilities we found:

\( P(W) = \left(\frac{4}{10}\right) \cdot \left(\frac{1}{3}\right) + \left(\frac{5}{10}\right) \cdot \left(\frac{2}{3}\right) \)

\( P(W) = \frac{4}{30} + \frac{10}{30} \)

\( P(W) = \frac{14}{30} = \frac{7}{15} \)

Now we have all the parts needed for Bayes' Theorem:

• \( P(W | B1) = \frac{4}{10} \)
• \( P(B1) = \frac{1}{3} \)
• \( P(W) = \frac{7}{15} \)

Substitute these values into the formula for \( P(B1 | W) \):

\( P(B1 | W) = \frac{\left(\frac{4}{10}\right) \cdot \left(\frac{1}{3}\right)}{\frac{7}{15}} \)

First, calculate the numerator:

\( \left(\frac{4}{10}\right) \cdot \left(\frac{1}{3}\right) = \frac{4 \cdot 1}{10 \cdot 3} = \frac{4}{30} = \frac{2}{15} \)

Now, divide the numerator by \( P(W) \):

\( P(B1 | W) = \frac{\frac{2}{15}}{\frac{7}{15}} \)

To divide fractions, multiply the numerator by the reciprocal of the denominator:

\( P(B1 | W) = \frac{2}{15} \cdot \frac{15}{7} \)

\( P(B1 | W) = \frac{2 \cdot 15}{15 \cdot 7} \)

Cancel out the 15s:

\( P(B1 | W) = \frac{2}{7} \)

The probability that the white ball was not drawn from Bag-2 (i.e., was drawn from Bag-1) is \( \frac{2}{7} \).

Let's summarize the probabilities:

Revision Table: Key Concepts in Probability

Additional Information on Probability with Bags and Die Rolls

Problems combining die rolls with drawing from bags are common in probability. The die roll (or coin flip, etc.) acts as an initial random event that determines the conditions for the second random event (drawing from a specific bag). The structure typically involves:

• Calculating the probabilities of the initial events (e.g., which bag is chosen).
• Calculating the conditional probabilities of the second event (e.g., drawing a specific color ball) given the outcome of the first event.
• Using the Law of Total Probability to find the overall probability of the second event.
• Using Bayes' Theorem to find the conditional probability of an initial event given the outcome of the second event.

In this specific problem, "not black" means "white". The condition is that a white ball is drawn. The event we are interested in is that Bag-1 was chosen. Therefore, it is a direct application of finding P(B1 | W).

It is crucial to correctly identify the probabilities P(Bag chosen) and P(Color | Bag chosen) from the problem description to accurately apply the probability formulas.

This type of problem demonstrates how probabilities from sequential or dependent events are calculated and how new information (the color of the ball drawn) updates our belief about a prior event (which bag was chosen).