Problem 6 - Algebra 2 Course in Mathematics for the IIT-JEE and Other Engineering Exams

Question

\(A, B\) and \(C\) are three horses participating in a race. The chance of \(A\) 's win is double of \(B\) and chance of \(B\) 's win is double of \(C\). Find out the probability for winning of each of them. Also find the probability that which horse wins the race, \(B\) or \(C\). [MP-2008]

Step-by-Step Solution

Answer

The probabilities are: P(C) = 1/7, P(B) = 2/7, P(A) = 4/7; and P(B or C) = 3/7.

Step 1: Define probabilities using variables

Let the probability of horse C winning be \( x \). Then, the probability of horse B winning is twice that of the probability of horse C, which is \( 2x \). Similarly, the probability of horse A winning is twice that of horse B, which is \( 4x \).

Step 2: Set up the total probability equation

Since one of the horses has to win the race, the sum of the probabilities must equal 1. Thus, we have:\[ x + 2x + 4x = 1 \]

Step 3: Solve for x

Combine the terms from the equation:\[ 7x = 1 \]Solve for \( x \):\[ x = \frac{1}{7} \]

Step 4: Find individual probabilities

Now that we know \( x = \frac{1}{7} \), calculate each probability:- Probability of C winning (\( P(C) \)) is \( x = \frac{1}{7} \).- Probability of B winning (\( P(B) \)) is \( 2x = \frac{2}{7} \).- Probability of A winning (\( P(A) \)) is \( 4x = \frac{4}{7} \).

Step 5: Calculate probability of B or C winning

We need the combined probability for either B or C winning, which is the sum \( P(B) + P(C) \):\[ P(B \text{ or } C) = \frac{2}{7} + \frac{1}{7} = \frac{3}{7} \]

Key Concepts

Probability Theory

Probability theory deals with understanding and calculating the likelihood of different events happening. In this context, we are interested in the probabilities of each horse winning the race.
Probability is expressed as a number between 0 and 1. A probability of 0 means the event will not happen, while a probability of 1 means it is certain to happen.
In our exercise, the probability of one horse winning does not affect the others because they are mutually exclusive events (only one can win). Therefore, we must ensure these probabilities sum to 1, covering all possible outcomes.
Using variables to define probabilities helps create equations that simplify the calculation. For example, if the probability of horse C winning is given as \(x\), and others depend on it, setting up a relationship helps us solve for one unknown to determine the others.

Problem Solving

Problem solving involves breaking down a question, using logic and mathematical operations to reach an answer. Let's discuss the steps in approaching the given problem.
The first step is identifying relationships among the variables. Here, probabilities have proportional relationships: C's probability is \(x\), and B's is \(2x\), making A's \(4x\).
This setup naturally leads to forming an equation based on the sum of their probabilities being 1, since one horse will win for sure:

• \( x + 2x + 4x = 1 \)

Solving such equations requires combining like terms to simplify the process:

• Adding the terms: \( 7x = 1 \)
• Then isolating \(x\): \( x = \frac{1}{7} \)

By systematically following these steps, the solution becomes clear and easy to follow, showcasing the importance of methodical problem-solving.

Algebraic Equations

Algebraic equations allow us to express mathematical relationships and solve for unknowns. In this exercise, all the given probabilities rely on the variable \(x\), creating a simple algebraic expression.
Starting with one variable makes it straightforward to relate all unknowns back to it. Here, after establishing \(x + 2x + 4x = 1\), each term corresponds to a horse's probability:

• C's probability is \( x = \frac{1}{7} \)
• B's probability is \( 2x = \frac{2}{7} \)
• A's probability is \( 4x = \frac{4}{7} \)

Algebra helps us systematically adjust any variable or expression to work with data. Solving for \(x\) first streamlined the process, enabling the cascading solution of other probabilities. Each part of the equation fits logically, ensuring the sum of parts equals the whole, which is 1 in this scenario.