Problem 104 - A Complete Resource Book in Mathematics for JEE Main

Question

A bag contains \(a\) white and \(b\) black balls. Two players \(A\) and \(B\) alternately draw a ball from the bag, replacing the ball each time after the draw till one of them draws a white ball and and wins the game. If \(A\) begins the game and the probability of \(A\) winning the game is three times that of \(B\), then \(a: b=\) (A) \(2: 1\) (B) \(3: 1\) (C) \(3: 2\) (D) none of these

Step-by-Step Solution

Answer

The ratio is \(3:1\). Therefore, the answer is (B) \(3: 1\).

Step 1: Define known values and probabilities

There are \(a\) white balls and \(b\) black balls. The probability that player A draws a white ball in one draw is \(\frac{a}{a+b}\), and similarly, for a black ball, it is \(\frac{b}{a+b}\). Player B has the same probabilities on their turn.

Step 2: Express probabilities of winning

Let the probability that player A wins the game be \(P(A)\), and the probability that player B wins be \(P(B)\). According to the problem, \(P(A) = 3\cdot P(B)\).

Step 3: Calculate recurrence relation for probability

The probability that A wins, \(P(A)\), is the sum of the probability of drawing a white ball on the first attempt (\(\frac{a}{a+b}\)) or drawing a black ball followed by B drawing a black ball, and then A winning (\(\frac{b}{a+b} \frac{b}{a+b} P(A)\)). Thus, \(P(A) = \frac{a}{a+b} + \frac{b^2}{(a+b)^2} P(A)\).

Step 4: Simplify equation for probability

Rearranging the equation gives \( P(A) \left( 1 - \frac{b^2}{(a+b)^2} \right) = \frac{a}{a+b} \). Then, solving for \(P(A)\), we get \( P(A) = \frac{a(a+b)^2}{a(a+b)^2 - b^2(a+b)} \).

Step 5: Express B's probability, use given condition

Using \(P(B) = 1 - P(A)\), we set up the equated probabilities \(P(A) = 3\times(1-P(A))\). Substituting \(X\) as \(P(A)\), \(X = 3(1-X)\).

Step 6: Solve for given condition on probabilities

Solve \( X = 3 - 3X \) to get \(4X = 3\), giving \(X = \frac{3}{4}\). Also, \(1-X = \frac{1}{4}\), so \(P(B) = \frac{1}{4}\).

Step 7: Derive a ratio for a:b

From the equation \( P(A) = \frac{a(a+b)^2}{a(a+b)^2 - b^2(a+b)} = \frac{3}{4} \), set \( P(B) = \frac{1}{4}\), then simplify this fraction to find the required relation \( \frac{a}{b} \). After equating and simplifying, this results in \(a:b = 3:1\).

Key Concepts

Conditional Probability

Conditional probability is an essential concept when analyzing outcomes that depend on previous events. In this exercise, each player's chance of drawing a white or black ball depends on the current contents of the bag, which remains constant due to the replacement of balls.
For player A, who starts the game, the initial conditional probability of winning directly is the chance of drawing a white ball:

• Probability of A drawing a white ball = \(\frac{a}{a+b}\)

This calculation assumes no previous draws because A is the first player to draw. If A draws a black ball, player B's future probabilities are conditioned on A’s initial draw.
This exercise leverages the idea that each draw and subsequent game state provides a new set of conditions, which shape the likelihoods for future actions. Both players' strategic advantage is contingent upon their respective probabilities of drawing a white, the ultimate winning draw, showcasing the role of conditional probability in game theory.

Probability Ratio

The probability ratio in this exercise breaks down the comparative likelihood of each player winning the game. As detailed, the victory chances of player A are three times those of player B, which directly leads to setting up the equation:

• \(P(A) = 3 \cdot P(B)\)

Understanding this ratio is crucial as it provides insight into the balance of fairness or bias within the game settings. By comparing probabilities, we see how much more likely it is for A to win over B, given that they both have equal chances on each turn.
The probability ratio helps students grasp comparative probability, which is common in real-world scenarios where outcomes depend on relative advantages rather than absolutes. Translating this ratio into a functioning equation facilitates deeper understanding of how different probabilities interact and affect each other within a bounded system like this repetitive game.

Recurrence Relation

Recurrence relations offer a mathematical way to define sequences or steps built from preceding outcomes. In this exercise, the recurrence relation helps find the equation for \(P(A)\), focusing on actions and their subsequent results in the game. The formula developed is:

• \(P(A) = \frac{a}{a+b} + \frac{b^2}{(a+b)^2} \cdot P(A)\)

Here, the recurrence relation sets up a loop—player A’s future draws are calculated based on whether they initially fail. This relation indicates that A can repeatedly try until a white ball is drawn or the losing condition persists.
Recurrence relations are essential in defining processes where a sequence of similar events take place, aiding in predictive modeling. In this case, solving the relations ultimately simplifies the probability evaluation, contributing to the understanding of how ongoing patterns impact final outcomes—becoming foundational in strategic assessments in probability-based games.