Problem 6 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

A box \(A^{\prime}\) contanis 2 white, 3 red and 2 black balls. Another box ' \(B\) ' contains 4 white, 2 red and 3 black balls. If two balls are drawn at random, without replacement, from a randomly selected box and one ball turns out to be white while the other ball turns out to be red, then the probability that both balls are drawn from box \({ }^{\prime} B^{\prime}\) is [Online April 15, 2018] (a) \(\frac{7}{16}\) (b) \(\frac{9}{32}\) (c) \(\frac{7}{8}\) (d) \(\frac{9}{16}\)

Step-by-Step Solution

Answer

The probability is \(\frac{9}{16}\) (Option d).

Step 1: Understand the Problem

We want to calculate the probability that two drawn balls, one white and one red, both come from box \(B\). First, identify the total possibilities for drawing from each box and the probability scenario for choosing box \(B\).

Step 2: Probability of Choosing Box A'

Box \(A'\) contains 2 white, 3 red, and 2 black balls, totaling 7 balls. Probability of choosing box \(A'\) is \(\frac{1}{2}\).

Step 3: Probability of Drawing White and Red from Box A'

For box \(A'\), the probability of drawing one white and one red is calculated by selecting 1 white from 2 and 1 red from 3. Using combinations, the probability is \(\frac{\binom{2}{1} \cdot \binom{3}{1}}{\binom{7}{2}} = \frac{2 \cdot 3}{21} = \frac{2}{7}\).

Step 4: Probability of Choosing Box B

Box \(B\) contains 4 white, 2 red, and 3 black balls, totaling 9 balls. Probability of choosing box \(B\) is \(\frac{1}{2}\).

Step 5: Probability of Drawing White and Red from Box B

For box \(B\), the probability of drawing one white and one red is calculated by selecting 1 white from 4 and 1 red from 2. Using combinations, the probability is \(\frac{\binom{4}{1} \cdot \binom{2}{1}}{\binom{9}{2}} = \frac{4 \cdot 2}{36} = \frac{2}{9}\).

Step 6: Total Probability of Drawing One White and One Red

Calculate the total probability for getting one white and one red by adding the probabilities of these events happening from both boxes. This is \(\frac{1}{2} \cdot \frac{2}{7} + \frac{1}{2} \cdot \frac{2}{9} = \frac{1}{7} + \frac{1}{9} = \frac{16}{63}\).

Step 7: Probability of Being from Box B Given Conditions

Apply Bayes' Theorem to find the probability that these two balls come from box \(B\) given that we drew one white and one red. This is \(\frac{\frac{1}{2} \cdot \frac{2}{9}}{\frac{16}{63}} = \frac{9}{16}\).

Key Concepts

Combinatorial Analysis

Combinatorial analysis helps us solve counting problems involving various objects. In this scenario, we're dealing with drawing balls from two boxes.
Each box contains balls of different colors, and we want to know the likelihood of selecting certain combinations of colors. This is where combinatorial analysis comes in.

To calculate these probabilities, we use combinations, not permutations, because the order of drawing doesn't matter.
The combination formula is given by \(\binom{n}{r} = \frac{n!}{r! \cdot (n-r)!}\), where \(n\) is the total number of items to choose from, and \(r\) is the number of items to choose.

• For example, when selecting one white ball and one red ball from Box \(A'\), we first calculate:\[ \binom{2}{1} \cdot \binom{3}{1} = 2 \cdot 3 = 6. \]
• We then divide by the total possible combinations of drawing two balls from Box \(A'\): \(\binom{7}{2} = 21.\)
• The probability becomes \(\frac{6}{21} = \frac{2}{7}.\)

This method applies similarly when calculating for Box \(B\). Combinatorics enables us to determine the likelihood of particular events without enumerating each possibility.

Bayes' Theorem

Bayes' Theorem is instrumental in solving problems involving conditional or dependent probabilities. It allows us to update our understanding of an event's probability based on new evidence.
In this exercise, we want to find the probability that both balls came from Box \(B\) given that we drew one white and one red ball.

Bayes' Theorem is articulated through the following formula: \[ P(A|B) = \frac{P(B|A) \cdot P(A)}{P(B)} \]
The theorem uses:

• \(P(A|B)\): Probability of the event \(A\) given event \(B\).
• \(P(B|A)\): Likelihood of event \(B\) given \(A\) has occurred.
• \(P(A)\) and \(P(B)\): The prior probabilities of events \(A\) and \(B\) occurring independently.

Applying it to the problem:

• Event \(A\) is drawing from Box \(B\).
• Event \(B\) is drawing one white and one red ball.

We calculated \(P(B) = \frac{16}{63}\) from total probabilities and \(P(B|A) = \frac{1}{2} \cdot \frac{2}{9}\).
Plugging into the formula, the result comes out as \(\frac{9}{16}\), which is the probability that both drawn balls came from Box \(B\) given the observed outcome.

Conditional Probability

Conditional probability is the likelihood of an event occurring given that another event has already happened.
Understanding this concept is key as it underlies much of both Bayes' Theorem and general probability tasks.

Conditional probability is denoted as \(P(A|B)\) and calculated by dividing the probability of both events occurring by the probability of the known event: \[ P(A|B) = \frac{P(A \cap B)}{P(B)} \]

• In this exercise, we seek \(P(\text{from Box } B |\text{white and red})\).
• Prior events are the drawing of one white and one red ball.

Previously calculated values are used within this formula:

• The joint probability of drawing one white and one red from Box \(B\) is \(\frac{1}{2} \times \frac{2}{9} = \frac{1}{9}\).
• The probability of drawing one white and one red from either box is \(\frac{16}{63}\).

By substituting these into the conditional probability formula, the probability that both balls came from Box \(B\) can be confirmed as \(\frac{9}{16}\). This showcases how conditional probability allows us to focus on specific scenarios, even within more complex, overlapping events.