Problem 11 - Algebra 2 Course in Mathematics for the IIT-JEE and Other Engineering Exams

Question

Coloured balls are distributed in three bags, as shown in the following table: $$ \begin{array}{lccc} \text { Bag } & \text { Black } & \text { White } & \text { Red } \\ \text { I } & 2 & 1 & 3 \\ \text { II } & 4 & 2 & 1 \\ \text { III } & 5 & 4 & 3 \end{array} $$ A bag is selected at random and then two balls are randomly drawn from the selected bag. They happen to be white and red. What is the probability that they came from bag II?

Step-by-Step Solution

Answer

The probability that the balls came from Bag II is \(\frac{22}{141}\).

Step 1: Identify Total Probability

First, calculate the probability of drawing two different colored balls, specifically one white and one red, from any given bag.In **Bag I**, the probability of drawing a white and red ball is calculated as follows:- Total combinations of drawing 2 balls from Bag I: \( \binom{6}{2} = 15 \)- Favorable combinations (drawing one white, one red): product of drawing 1 white from 1 white ball and 1 red from 3 red balls: \( \binom{1}{1} \cdot \binom{3}{1} = 3 \)- Probability: \( \frac{3}{15} = \frac{1}{5} \).In **Bag II**, the probability is:- Total combinations of drawing 2 balls from Bag II: \( \binom{7}{2} = 21 \)- Favorable combinations: \( \binom{2}{1} \cdot \binom{1}{1} = 2 \)- Probability: \( \frac{2}{21} \).In **Bag III**, the probability is:- Total combinations of drawing 2 balls from Bag III: \( \binom{12}{2} = 66 \)- Favorable combinations: \( \binom{4}{1} \cdot \binom{3}{1} = 12 \)- Probability: \( \frac{12}{66} = \frac{2}{11} \).Each bag is chosen with equal probability, \( \frac{1}{3} \).

Step 2: Apply Bayes' Theorem

Now, use Bayes' Theorem to find the probability that the selected bag was Bag II, given the event that one white and one red ball were drawn.Bayes' Theorem is given by:\[P(B_2 \mid E) = \frac{P(E \mid B_2) \cdot P(B_2)}{P(E)}\]Where:- \( P(B_2) \) is the probability of selecting Bag II: \( \frac{1}{3} \).- \( P(E \mid B_2) \) is the probability of drawing a white and red ball from Bag II: \( \frac{2}{21} \).- \( P(E) \) is the total probability of drawing a white and red ball, which includes the contributions from all bags.

Step 3: Calculate Total Probability for White and Red

To calculate \( P(E) \), find the weighted probability of drawing one white and one red ball from any bag:\[P(E) = P(E \mid B_1) \cdot P(B_1) + P(E \mid B_2) \cdot P(B_2) + P(E \mid B_3) \cdot P(B_3)\]where each \( P(B_i) = \frac{1}{3} \).Thus,\[P(E) = \frac{1}{5} \cdot \frac{1}{3} + \frac{2}{21} \cdot \frac{1}{3} + \frac{2}{11} \cdot \frac{1}{3}\]Simplifying,\[P(E) = \frac{1}{15} + \frac{2}{63} + \frac{2}{33} \]Convert to a common denominator and add:\[P(E) = \frac{77}{693} + \frac{22}{693} + \frac{42}{693} = \frac{141}{693}\]Simplify to \( \frac{47}{231} \).

Step 4: Compute Final Probability

Use the computed probabilities to find \( P(B_2 \mid E) \):\[P(B_2 \mid E) = \frac{\frac{2}{21} \cdot \frac{1}{3}}{\frac{47}{231}}\]Simplifying,\[P(B_2 \mid E) = \frac{2}{21} \cdot \frac{1}{3} \cdot \frac{231}{47} = \frac{2 \times 11}{47}\]Which simplifies to \(\frac{22}{141}\).

Step 5: Result

The probability that the balls came from Bag II is \(\frac{22}{141}\).

Key Concepts

Understanding Probability Theory

Probability theory is a branch of mathematics concerned with the analysis of random events and the likelihood of different outcomes. It helps us measure uncertainty and model random processes.
In simple terms, probability gives us the ability to quantify how likely something is to happen. It ranges from 0 to 1, where 0 indicates impossibility, and 1 indicates certainty.
When solving probability problems, we often deal with events and their associated probabilities.Some key terms in probability theory include:

• Experiment: A process or action, such as tossing a die or drawing a card, with a well-defined set of outcomes.
• Outcome: A possible result of an experiment.
• Event: A combination of outcomes.
• Sample Space: All possible outcomes of an experiment, denoted as \( S \).
• Probability of an event \( A \): Denoted as \( P(A) \), it measures the likelihood of event \( A \) occurring, calculated as the ratio of favorable outcomes for \( A \) to the total number of possible outcomes.

Probability theory is foundational for understanding more advanced concepts such as combinatorics and conditional probability, which are crucial in complex problems like the one in our exercise.

Basics of Combinatorics

Combinatorics is a field of mathematics focused on counting and arranging objects. It's essential in probability, as it helps us determine the number of ways an event can occur.
Consider drawing balls from a bag, where we need to count the combinations of different colored balls without regard to order.In our exercise, combinatorics was used to calculate different possible combinations for drawing balls:

• Combinations: The selection of items from a group, without considering the order. Represented using the symbol \( \binom{n}{r} \), where \( n \) is the total number of items, and \( r \) is the number of items to choose. It is calculated as \( \binom{n}{r} = \frac{n!}{r! (n-r)!} \).
• Applications: In the exercise, combinations were used to find the total ways of drawing two balls from each bag and the favorable combinations of a specific color pairing.

Combinatorics simplifies the process of determining event probabilities by providing a precise method to count outcomes. It lays the groundwork for evaluating how likely it is for a certain event to occur in probability theory.

Understanding Conditional Probability

Conditional probability is the likelihood of an event occurring, given that another event has already happened. It's a crucial concept when we want to update our predictions based on new information.
The conditional probability of event \( A \) given event \( B \) is denoted as \( P(A \mid B) \) and can be calculated as:\[P(A \mid B) = \frac{P(A \cap B)}{P(B)}\]where \( P(A \cap B) \) is the probability that both events \( A \) and \( B \) occur together.In the exercise, Bayes' Theorem comes into play, which is a powerful tool for finding conditional probabilities:

• Bayes' Theorem: It's used to update the probability of a hypothesis or event based on new evidence or information.
• Formula: \[ P(B_2 \mid E) = \frac{P(E \mid B_2) \cdot P(B_2)}{P(E)} \]

Bayes' Theorem effectively helped determine the probability that the balls came from Bag II, given the draw consisted of one white and one red ball. In real life, conditional probability is applied in fields like medicine, finance, and engineering to make informed decisions based on probabilities.