Problem 157 - A Complete Resource Book in Mathematics for JEE Main

Question

Three houses are available in a locality. Three persons apply for the houses. Each applies for one house without consulting others. The probability that all the three apply for the same house is [2005] (A) \(\frac{2}{9}\) (B) \(\frac{1}{9}\) (C) \(\frac{8}{9}\) (D) \(\frac{7}{9}\)

Step-by-Step Solution

Answer

The probability is \( \frac{1}{9} \).

Step 1: Understanding the Problem

We need to find the probability that all three people apply for the same house among the three available houses. This involves looking at the combinations where all choose the same house and then comparing to the total possible choices.

Step 2: Total Number of Combinations

Each of the three people can choose any of the three houses independently. Therefore, each person has 3 choices, which gives a total number of combinations as: \[ 3 \times 3 \times 3 = 27 \]

Step 3: Combinations for All Applying to the Same House

For all three people to apply for the same house, they must all choose the same one of the three houses. There are 3 houses to pick from, so the number of favorable outcomes is the same as the number of houses: \[ 3 \]

Step 4: Calculate the Probability

The probability that all three people apply to the same house is the number of favorable outcomes divided by the total number of outcomes. Thus, the probability is: \[ \frac{3}{27} = \frac{1}{9} \]

Step 5: Verify the Answer

Verify the calculation by checking if the logic holds: Each person independently has 3 choices, totaling 27 possible ways they can apply. Of those, only 3 combinations work for this particular scenario where all apply to the same house.

Key Concepts

Combinatorics

Combinatorics is the area of mathematics that deals with counting, arrangement, and combination of objects. It is especially useful when determining the number of possible outcomes in a given situation. In the problem of three people applying to three different houses, combinatorics helps us understand the number of possible ways this can happen.

When each of the three people has 3 houses to choose from, you multiply the number of possibilities for each person to find the total number of combinations. Thus, you calculate it as follows:

• Person 1: 3 choices
• Person 2: 3 choices
• Person 3: 3 choices

By multiplication, you find that there are \(3 \times 3 \times 3 = 27\) potential ways all three can select houses. This multiplication is known as the "fundamental principle of counting." It's a foundational concept in combinatorics, allowing us to count the comprehensive number of ways an event can unfold given independent options.

Independent Events

Independent events are scenarios where the outcome of one event does not affect the outcome of another. This is crucial in probability, making sure the calculation of outcomes accurately reflects the nature of the events.

In this exercise, each person's choice of house does not depend on what the other two people choose. Let’s break down how this applies:

• Person A chooses a house.
• Person B's choice does not change regardless of Person A's choice.
• Person C's choice remains unaffected by the selections of Persons A and B.

This independence is what allows us to use simple multiplication to find the total number of outcomes, since each person's choice is a separate and independent event in probability theory.

Favorable Outcomes

Favorable outcomes are the specific outcomes within all possible scenarios that satisfy the conditions of the event we're interested in. In probability, analyzing favorable outcomes helps determine how likely an event will occur.

In this problem, the favorable outcome is the scenario where all three people apply for the same house. With 3 houses available, they could all choose house 1, house 2, or house 3. Hence, there are 3 favorable outcomes.

Calculating probability involves dividing the number of favorable outcomes by the total number of possible outcomes. Thus, the probability that all three apply to the same house is calculated as shown:

• Total outcomes = 27 (from combinatorics)
• Favorable outcomes = 3 (one for each house)

The probability that everyone applies to the same house becomes:\[\frac{3}{27} = \frac{1}{9}\]Understanding favorable outcomes is key to solving probability problems accurately, ensuring you're focusing on the correct subset of all possibilities.