Problem 12 - Algebra 2 Course in Mathematics for the IIT-JEE and Other Engineering Exams

Question

\(A\) and \(B\) throw a pair of dice. \(A\) wins if he throws 6 before \(B\) throws 7 and \(B\) wins if hethrows 7 before \(A\) throws 6 . If \(A\) begins, what is his chance of winning? (a) \(30 / 61\) [MNR 1995] (c) \(61 / 30\) (1) olution (a) Let \(E_{1}\) denote the event of \(A\) 's throwing 6 and \(E_{2}\) the event of \(B\) 's throwing 7 with a pair of dice. Then \(\bar{E}_{1}, \bar{E}_{2}\) are the complementary events. There are five ways of obtaining 6, namely, \((1,5),(2,4),(3,3),(4,2),(5,1)\) and similarly there are six ways of getting 7 , namely, \((1,6)\), \((2,5),(3,4),(4,3),(5,2),(6,1)\) $$ \therefore P\left(E_{1}\right)=\frac{5}{36} $$ and \(P\left(\bar{E}_{1}\right)=1-\frac{5}{36}=\frac{31}{36}\) $$ P\left(E_{2}\right)=\frac{6}{36}=\frac{1}{6} $$ and \(P\left(\bar{E}_{2}\right)=1-\frac{1}{6}=\frac{5}{6}\) It is given that \(A\) starts the game and he will win in the following mutually exclusive ways. (i) \(E_{1}\) happens, i.e., \(A\) wins at the first draw. (ii) \(\vec{E}_{1} \cap \bar{E}_{2} \cap E_{1}\) happens, i.e., \(A\) wins at the third draw when both \(A\) and \(B\) fail at \(1^{s t}\) and \(2^{\text {nd }}\) draw. (iii) \(\bar{E}_{1} \cap \bar{E}_{2} \cap \bar{E}_{1} \cap \bar{E}_{2} \cap E_{1}\) happens, i.e., \(A\) wins at the \(5^{\text {th }}\) draw when both \(A\) and \(B\) fail at \(1^{\text {st }}, 2^{\text {nd }}, 3^{\text {rd }}\) and \(4^{\text {th }}\) draw and so on \(\ldots\) Hence the required probability of \(A\) winning say \(P(A)\) is given by \(P(A)=P(\mathrm{i})+P(\mathrm{ii})+P(\mathrm{iii})\) \(+\ldots\)throws 7 before \(A\) throws 6 . If \(A\) begins, what is his chance of winning? (a) \(30 / 61\) [MNR 1995] (c) \(61 / 30\) (1) olution (a) Let \(E_{1}\) denote the event of \(A\) 's throwing 6 and \(E_{2}\) the event of \(B\) 's throwing 7 with a pair of dice. Then \(\bar{E}_{1}, \bar{E}_{2}\) are the complementary events. There are five ways of obtaining 6, namely, \((1,5),(2,4),(3,3),(4,2),(5,1)\) and similarly there are six ways of getting 7 , namely, \((1,6)\), \((2,5),(3,4),(4,3),(5,2),(6,1)\) $$ \therefore P\left(E_{1}\right)=\frac{5}{36} $$ and \(P\left(\bar{E}_{1}\right)=1-\frac{5}{36}=\frac{31}{36}\) $$ P\left(E_{2}\right)=\frac{6}{36}=\frac{1}{6} $$ and \(P\left(\bar{E}_{2}\right)=1-\frac{1}{6}=\frac{5}{6}\) It is given that \(A\) starts the game and he will win in the following mutually exclusive ways. (i) \(E_{1}\) happens, i.e., \(A\) wins at the first draw. (ii) \(\vec{E}_{1} \cap \bar{E}_{2} \cap E_{1}\) happens, i.e., \(A\) wins at the third draw when both \(A\) and \(B\) fail at \(1^{s t}\) and \(2^{\text {nd }}\) draw. (iii) \(\bar{E}_{1} \cap \bar{E}_{2} \cap \bar{E}_{1} \cap \bar{E}_{2} \cap E_{1}\) happens, i.e., \(A\) wins at the \(5^{\text {th }}\) draw when both \(A\) and \(B\) fail at \(1^{\text {st }}, 2^{\text {nd }}, 3^{\text {rd }}\) and \(4^{\text {th }}\) draw and so on \(\ldots\) Hence the required probability of \(A\) winning say \(P(A)\) is given by \(P(A)=P(\mathrm{i})+P(\mathrm{ii})+P(\mathrm{iii})\) \(+\ldots\)

Step-by-Step Solution

Answer

A's chance of winning is \(\frac{30}{61}\).

Step 1: Understand the Problem Statement

We need to find the probability that player \(A\) wins the game. \(A\) wins if he throws a 6 before \(B\) throws a 7, and \(B\) wins if he throws a 7 before \(A\) throws a 6. \(A\) starts the game.

Step 2: Probability of Throwing Specific Numbers

The probability that \(A\) throws a 6 is given by the number of combinations that sum to 6 (which is 5) divided by the total number of combinations with two dice (36). So, \(P(E_1) = \frac{5}{36}\) and \(P(\bar{E}_1) = 1 - \frac{5}{36} = \frac{31}{36}\). Likewise, the probability that \(B\) throws a 7 is 6 combinations out of 36, so \(P(E_2) = \frac{6}{36} = \frac{1}{6}\) and \(P(\bar{E}_2) = 1 - \frac{1}{6} = \frac{5}{6}\).

Step 3: Enumerate Winning Scenarios for A

\(A\) can win in several mutually exclusive scenarios: (i) \(A\) throws a 6 on his first throw, (ii) both \(A\) and \(B\) fail to get the required numbers in their first turns, and \(A\) gets a 6 on his second turn, and so on. These scenarios involve alternate failures by \(A\) and \(B\) until \(A\) succeeds.

Step 4: Calculate the Probability of Each Scenario

Let's compute the probabilities for some initial scenarios: \(P(\text{i}) = P(E_1) = \frac{5}{36}\), \(P(\text{ii}) = P(\bar{E}_1) \cdot P(\bar{E}_2) \cdot P(E_1) = \frac{31}{36} \cdot \frac{5}{6} \cdot \frac{5}{36}\), and similarly for other patterns involving \(\bar{E}_1\) and \(\bar{E}_2\).

Step 5: Use Infinite Series to Sum All Scenarios

The probability of \(A\) winning is the sum of the probabilities of all scenarios: \[P(A) = \frac{5}{36} + \frac{31}{36} \cdot \frac{5}{6} \cdot \frac{5}{36} + \left(\frac{31}{36} \cdot \frac{5}{6}\right)^2 \cdot \frac{5}{36} + \ldots \] This is an infinite geometric series with first term \(a = \frac{5}{36}\) and common ratio \(r = \frac{31}{36} \cdot \frac{5}{6}\).

Step 6: Solve the Geometric Series

The sum \(S\) of an infinite geometric series can be computed using the formula: \[ S = \frac{a}{1 - r} \] where \(a = \frac{5}{36}\) and \(r = \frac{31}{36} \cdot \frac{5}{6} = \frac{155}{216}\). Substituting these values, \(S = \frac{5/36}{1 - 155/216} = \frac{5/36}{61/216}\), which simplifies to \(\frac{5}{36} \times \frac{216}{61} = \frac{30}{61}\).

Key Concepts

Geometric Series

A geometric series is a mathematical concept where each term in the series is a fixed multiple of the previous term. In the context of probability, geometric series are often used to sum probabilities of infinite events that follow a pattern.
For example, if you have a first term "a" and a common ratio "r", the sum of an infinite geometric series can be calculated using the formula: \( S = \frac{a}{1 - r} \). This formula applies when \(|r| < 1\), ensuring that the series converges to a finite sum.
In our problem, player A’s probability of winning involves repeating cycles of events (A failing, B failing, then A succeeding eventually). The common ratio, "r", encapsulates the cycle of failures before a success. As you can see, breaking probabilities into a sequence of geometric patterns helps simplify complex infinite outcomes.

Mutually Exclusive Events

Mutually exclusive events are situations in which the occurrence of one event implies the non-occurrence of another. This means both events cannot happen at the same time.
In probabilistic terms, if events A and B are mutually exclusive, then \( P(A \cap B) = 0 \). For instance, in the problem with players A and B, the scenarios where A wins are mutually exclusive paths, such as winning immediately on the first roll or after one cycle of misses.
Each scenario is independent of the others, allowing us to add their probabilities directly to find the total probability of A winning. Understanding mutually exclusive events helps us isolate and calculate distinct paths leading to success in these probability games.

Combinatorics

Combinatorics is the study of counting, arrangement, and combination of objects. It finds a robust application in defining probabilities, especially when dealing with outcomes of events, like rolling dice.
In our dice-throwing scenario, we use combinatorics to determine the number of ways to achieve specific sums on a pair of dice. For example, there are 5 ways to sum to 6: \((1,5), (2,4), (3,3), (4,2), (5,1)\), and 6 ways to sum to 7: \((1,6), (2,5), (3,4), (4,3), (5,2), (6,1)\).
By calculating these combinations, we can derive the probabilities necessary for each event, providing a foundational tool for solving probability problems.

Probability of Complementary Events

Complementary events consist of all possible outcomes of an event that are not in the original event. If "E" is an event, then the complementary event "\(\bar{E}\)" includes all outcomes where \(E\) does not happen.
The probability of complementary events helps fill in gaps when calculating complex probabilities by making use of the rule: \( P(\bar{E}) = 1 - P(E) \).
In our dice game, if \(E_1\) represents A throwing a 6, then \(\bar{E}_1\) is the event where A does not throw a 6. This complementary probability allows us to factor in misses and repeated turns until success, which is integral in setting up the geometric series for A’s continued play logic.