Problem 30 - Algebra 2 Course in Mathematics for the IIT-JEE and Other Engineering Exams

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 (a) \(8 / 9\) (b) \(7 / 9\) (c) \(2 / 9\) (d) \(1 / 9\)

Step-by-Step Solution

Answer

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

Step 1: Identify Total Outcomes

Each of the three applicants can choose from any of the three houses. Thus, for each applicant, there are 3 choices available. This means the total number of ways all three applicants can apply for the houses is calculated as follows: \(3 \times 3 \times 3 = 27\) total possible outcomes.

Step 2: Determine Favorable Outcomes

The favorable outcome is that all three applicants apply for the same house. For this to occur, they can either all choose house 1, house 2, or house 3. Therefore, there are 3 favorable outcomes (each representing the choice of one house by all three applicants).

Step 3: Calculate Probability

The probability of an event is calculated by dividing the number of favorable outcomes by the total number of possible outcomes. Here the probability is given by:\[\text{Probability} = \frac{\text{Number of Favorable Outcomes}}{\text{Total Number of Outcomes}} = \frac{3}{27} = \frac{1}{9}\]

Key Concepts

Total Outcomes

In probability, understanding the concept of total outcomes is crucial. Total outcomes refer to all possible ways an event can happen. In our exercise, this involves three individuals applying for houses independently. Each person has a choice of 3 houses.
To calculate the total number of outcomes, imagine each individual making a decision. The first person has 3 choices, the second person also has 3 choices, and the third person again has 3. These choices compound based on the number of individuals, leading us to multiply the choices together. Thus, the total number of outcomes is calculated as:

• 3 choices for the first individual
• 3 choices for the second individual
• 3 choices for the third individual

Using multiplication, this gives us a total of \(3 \times 3 \times 3 = 27\) possible ways the applications can be distributed. This means there are 27 different combinations of house assignments amongst the applicants.

Favorable Outcomes

Once we determine the total outcomes, we focus on the favorable outcomes. Favorable outcomes are those that meet the event criteria we're interested in.
In this case, the criterion is that all applicants apply for the same house. This means for each specific house, all three applicants must choose it together.
To find the number of favorable outcomes, consider:

• All three applicants choose house 1
• All three applicants choose house 2
• All three applicants choose house 3

There are 3 distinct scenarios where all applicants opt for the same house. Thus, the number of favorable outcomes is 3. Understanding favorable outcomes helps in determining the likelihood of a specific event which leads us to the next step, calculating probability.

Probability Calculation

Calculating the probability of an event involves dividing the number of favorable outcomes by the total outcomes. This ratio gives us an insight into how likely an event is to occur.
For our exercise, the probability can be calculated as follows: - We have 3 favorable outcomes (all applicants choose the same house) - We discovered there are 27 total possible outcomes in total With these numbers, the probability that all applicants pick the same house is: \[\text{Probability} = \frac{\text{Number of Favorable Outcomes}}{\text{Total Number of Outcomes}} = \frac{3}{27}\]
When simplified, this fraction becomes \(\frac{1}{9}\). Hence, the probability that all three individuals apply for the same house is \(\frac{1}{9}\). This method of calculation is foundational in probability, providing a precise measurement of an event's likelihood.