Problem 23 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

A bag contains 4 red and 6 black balls. A ball is drawn at random from the bag, its colour is observed and this ball along with two additional balls of the same colour are returned to the bag. If now a ball is drawn at random from the bag, then the probability that this drawn ball is red, is : \([2018]\) (a) \(\frac{2}{5}\) (b) \(\frac{1}{5}\) (c) \(\frac{3}{4}\) (d) \(\frac{3}{10}\)

Step-by-Step Solution

Answer

The probability is \(\frac{13}{30}\).

Step 1: Calculate Initial Probabilities

Calculate the initial probability of drawing a red ball from the bag. There are 4 red balls and 6 black balls, making a total of 10 balls in the bag. Thus, the probability of drawing a red ball is \(\frac{4}{10}\) or \(\frac{2}{5}\), and the probability of drawing a black ball is \(\frac{6}{10}\) or \(\frac{3}{5}\).

Step 2: Adding Balls Based on Initial Draw

If a red ball is drawn first, then we add two more red balls. If a black ball is drawn first, then we add two more black balls. This alters the total and the composition of the balls in the bag.

Step 3: Calculate Outcomes for Red Ball First

If a red ball is drawn first, the number of red balls becomes 4 + 3 = 7, and the total number of balls is 10 + 2 = 12. The probability of drawing the second ball as red given the first was red is \(\frac{7}{12}\).

Step 4: Calculate Outcomes for Black Ball First

If a black ball is drawn initially, the number of black balls becomes 6 + 3 = 9, and the total number of balls is 10 + 2 = 12. The probability of drawing the second ball as red given the first was black remains \(\frac{4}{12}\) or \(\frac{1}{3}\).

Step 5: Calculate the Overall Probability

To find the total probability of the second ball being red, use the law of total probability: \(P(\text{Red}) = P(\text{Red|Red first}) \cdot P(\text{Red first}) + P(\text{Red|Black first}) \cdot P(\text{Black first}) = \frac{7}{12} \times \frac{2}{5} + \frac{1}{3} \times \frac{3}{5}\).

Step 6: Perform the Calculations

Calculate each part of the expression: \(\frac{7}{12} \times \frac{2}{5} = \frac{14}{60} = \frac{7}{30}\) and \(\frac{1}{3} \times \frac{3}{5} = \frac{3}{15} = \frac{1}{5}\). Add these results to find the total probability: \(\frac{7}{30} + \frac{1}{5} = \frac{7}{30} + \frac{6}{30} = \frac{13}{30}\).

Step 7: Simplify the Answer if Possible

The calculated probability is \(\frac{13}{30}\), which is already in its simplest form.

Key Concepts

Conditional Probability

Conditional probability helps us understand the likelihood of an event occurring based on the occurrence of a related event.
This concept is crucial when the conditions of the environment influence the outcomes, as seen in this exercise. Let's explore how conditional probability plays a role in our ball drawing scenario.

In this exercise, there's a second draw of the ball, and its outcome relies on what happened in the first draw. If the first draw is a red ball, two additional red balls are added to the bag. The probability of drawing a red ball on the second draw depends on this previous event.

Using conditional probability, we first find the probability of drawing a red ball given that a red one was drawn initially:

• If the first ball is red, the probability of the second being red is given by: \( P(\text{Red second | Red first}) = \frac{7}{12} \).
• If the first ball is black, the probability of the second ball being red is calculated as: \( P(\text{Red second | Black first}) = \frac{1}{3} \).

Conditional probability allows us to adjust our expectations based on known conditions of earlier events. This helps in determining the overall chance of subsequent events based on initial outcomes.

Total Probability Theorem

The Total Probability Theorem allows us to find the overall probability of an event by combining different possible scenarios that lead to that event.
For our exercise, this theorem steps in to provide an overall probability for drawing a red ball on the second draw regardless of what happened in the first draw.

We start by calculating probabilities for each possible initial condition and how they affect the outcome. We use these individual probabilities along with the theorem to find the total probability of a second red ball:

• Probability of drawing a red ball first: \( P(\text{Red first}) = \frac{2}{5} \)
• Probability of drawing a black ball first: \( P(\text{Black first}) = \frac{3}{5} \)

Using these probabilities, apply the Total Probability Theorem:\[P(\text{Red}) = P(\text{Red|Red first}) \cdot P(\text{Red first}) + P(\text{Red|Black first}) \cdot P(\text{Black first})\]
The calculations show:

• \( \frac{7}{12} \times \frac{2}{5} = \frac{7}{30} \)
• \( \frac{1}{3} \times \frac{3}{5} = \frac{1}{5} = \frac{6}{30} \)

Add these results to find the total probability: \( \frac{7}{30} + \frac{6}{30} = \frac{13}{30} \).
This theorem helps to effectively pool together probabilities of different possible scenarios, calculating a cumulative probability of an event.

Combinatorics

Combinatorics, a field of mathematics concerning the counting, arrangement, and combination of objects, is crucial in many probability problems.
Although not directly used in the exercise calculations, it underpins the logic of selecting and adding balls in different scenarios.

In our problem, combinatorial insights help to nicely organize how the number of balls changes based on initial draws and how different events can combine. Here’s how:

• The initial selection of a ball can be seen as a combinatorial choice from the first ten balls, represented by \( \binom{10}{1} \).
• When analyzing outcomes, we consider new combinations - by adding two additional balls into the mix, altering how future selections can be made.

Understanding combinatorics isn't just about counting but about visualizing how those counts can change based on processes or events.
This aspect lays groundwork in understanding basic probability by tracking how different outcomes can manifest as drawn from the arrangement of items available.