Problem 3 - Algebra 2 Course in Mathematics for the IIT-JEE and Other Engineering Exams

Question

A bag contains 4 white and 5 black balls and another bag contains 3 white and 4 black balls. A ball is taken out from the first bag and without seeing its colour is put in the second bag. A ball is taken out from another bag. Find the probability that the ball drawn is white. [CBSE-94]

Step-by-Step Solution

Answer

The probability that the drawn ball is white is \( \frac{31}{72} \).

Step 1: Determine Initial Probabilities in Bag 1

Bag 1 has a total of 9 balls (4 white and 5 black). The probability of drawing a white ball from Bag 1 is given by the number of white balls divided by the total number of balls: \( P(W_1) = \frac{4}{9} \). Similarly, the probability of drawing a black ball from Bag 1 is \( P(B_1) = \frac{5}{9} \).

Step 2: Determine Outcomes after Transfer

If a white ball is transferred from Bag 1, Bag 2 will then have 4 white balls and 4 black balls. If a black ball is transferred, Bag 2 will have 3 white balls and 5 black balls.

Step 3: Calculate Probability of Drawing a White Ball after Transfer

The probability of drawing a white ball from Bag 2 given a white ball was transferred from Bag 1 is \( P(W_2 | W_1) = \frac{4}{8} = \frac{1}{2} \). The probability of drawing a white ball from Bag 2 given a black ball was transferred is \( P(W_2 | B_1) = \frac{3}{8} \).

Step 4: Apply the Law of Total Probability

Using the law of total probability, the overall probability of drawing a white ball from Bag 2 is \( P(W_2) = P(W_2 | W_1) \times P(W_1) + P(W_2 | B_1) \times P(B_1) \).\[ P(W_2) = \left( \frac{1}{2} \right) \times \left( \frac{4}{9} \right) + \left( \frac{3}{8} \right) \times \left( \frac{5}{9} \right) \]

Step 5: Simplify the Expression

Calculate each part:- \( \frac{1}{2} \times \frac{4}{9} = \frac{2}{9} \)- \( \frac{3}{8} \times \frac{5}{9} = \frac{15}{72} = \frac{5}{24} \)Sum these to find \( P(W_2) \):\[ P(W_2) = \frac{2}{9} + \frac{5}{24} \]Convert \( \frac{2}{9} \) to an equivalent fraction with a common denominator of 72: \( \frac{16}{72} \).\[ P(W_2) = \frac{16}{72} + \frac{15}{72} = \frac{31}{72} \]

Key Concepts

Law of Total Probability

The Law of Total Probability is a useful principle when you want to find the total probability of an event based on several different events that can lead to the same outcome. In this problem, we're interested in finding the probability of drawing a white ball from the second bag. First, we need to account for all the possible ways this can happen.
Imagine you have two possible scenarios when transferring a ball from Bag 1 to Bag 2, each affecting the contents of Bag 2 differently. This results in:

• Scenario 1: A white ball is transferred, leading to a mix of 4 white and 4 black balls in Bag 2.
• Scenario 2: A black ball is transferred, resulting in 3 white and 5 black balls in Bag 2.

The Law of Total Probability combines these scenarios, taking into account the initial probabilities from Bag 1 and conditional probabilities from Bag 2, ensuring every possibility is included in the final calculation. Thus, the formula\[ P(W_2) = P(W_2 | W_1) \times P(W_1) + P(W_2 | B_1) \times P(B_1) \]represents summing up all these products to get the overall probability of drawing a white ball from Bag 2.

Conditional Probability

Conditional Probability helps determine the probability of an event given that another event has occurred. It's about understanding how the likelihood of one event could change based on the occurrence of a related event.
In the context of this problem, conditional probability is used to describe what happens to Bag 2 after transferring a ball from Bag 1. There are specific probabilities related to whether a white or black ball was transferred:

• If a white ball is transferred to Bag 2, the probability of drawing a white ball from Bag 2 becomes \( P(W_2 | W_1) = \frac{1}{2} \).
• Alternatively, if a black ball is transferred, the changed probability of drawing a white ball from Bag 2 is \( P(W_2 | B_1) = \frac{3}{8} \).

Conditional probability is a critical part of solving this problem because it shows how the state of Bag 2 is affected directly by the transferred ball’s type, altering the subsequent probability of drawing a white ball.

Probability of Events

Understanding basic probability concepts is essential to solving problems like this one. Probability measures how likely an event is going to occur on a scale from 0 (impossible) to 1 (certain). In this problem, the main events are the type of ball being picked from the bags and the probability associated with these events.
The initial probabilities in Bag 1 calculate as follows:

• The probability of picking a white ball is \( P(W_1) = \frac{4}{9} \),
• and the probability of picking a black ball is \( P(B_1) = \frac{5}{9} \).

These probabilities are essential building blocks for determining what happens next, post-transfer.

Every decision in this sequence depends on calculating these probabilities step by step, assembling them through principles like the Law of Total Probability to derive a meaningful outcome. In this way, understanding each component of probability not only clarifies specific events but also supports solving the broader picture of the problem.