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Chapter 4: Problem 30

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) \(1 / 9\)

Short Answer

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

Step by step solution

01

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.
02

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).
03

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:\[\text{Probability} = \frac{\text{Number of Favorable Outcomes}}{\text{Total Number of Outcomes}} = \frac{3}{27} = \frac{1}{9}\]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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.

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Most popular questions from this chapter

In a combat between \(A, B\) and \(C, A\) tries to hit \(B\) and \(C\), and \(B\) and \(C\) try to hit \(A\). Probability of \(A, B\) and \(C\) hitting the targets are \(2 / 3,1 / 2\) and \(1 / 3\), respectively. If \(A\) is hit, find the probability that \(B\) hits \(A\) and \(C\) does not. (a) \(1 / 1\) (b) \(1 / 2\) (c) \(2 / 2\) (d) \(2 / 1\) (c) We have to find the probability of \(A\) being hit by \(B\) but not by \(C\), i.e.,$$ \begin{aligned} P\left(B C^{\prime} / A\right)=& \frac{P\left(A / B C^{\prime}\right) P\left(B C^{\prime}\right)}{P\left(A / B C^{\prime}\right) P\left(B C^{\prime}\right)+P\left(A B C^{\prime}\right) P\left(B^{\prime} C\right)} \\ &+P(A / B C) P(B C)+P\left(A / B^{\prime} C^{\prime}\right) P\left(R^{\prime} C^{\prime}\right) \end{aligned} $$ Now putting the values from the given data, we have

A man alternately tosses a coin and throws a dice beginning with the coin. The probability that he gets a head in the coin before he gets a 5 or 6 in the dice is (a) \(3 / 4\) (b) \(1 / 2\) (c) \(1 / 3\) (d) None

A random variable \(X\) has the probability distribution: \(\begin{array}{lcccccccc}X & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 \\ P(X) & 0.15 & 0.23 & 0.12 & 0.10 & 0.20 & 0.08 & 0.07 & 0.05\end{array}\) For the events \(E=\\{X\) is a prime number \(\\}\) and \(F=\\{X<4\\}\), then \(P(E \cup F)\) is: (a) \(0.77\) (b) \(0.87\) (c) \(0.35\) (d) \(0.50\)

In a test an examinee either guesses or copies or knows the answer to a multiple choice question with four choices. The probability that he makes a guess is \(1 / 3\) and the probability that he copies the answer is \(1 / 6\). The probabilitythat his answer is correct given that he copied it, is \(1 / 8\). The probability that he knew the answer to the question given that he correctly answered it is [IIT-1991] (a) \(24 / 29\) (b) \(25 / 24\) (c) \(29 / 24\) (d) \(24 / 25\) Solution (a) Let \(A_{1}\) be the event that the examinee guesses the answer; \(A_{2}\) the event that he copies the answer and \(A_{3}\) the event that he knows the answer. Also let \(A\) be the event that he answers correctly. Then as given, we have $$ P\left(A_{1}\right)=\frac{1}{3}, P\left(A_{2}\right)=\frac{1}{6}, P\left(A_{3}\right)=1-\frac{1}{3}-\frac{1}{6}=\frac{1}{2} $$ (We have assumed here that the events \(A_{1}, A_{2}\) and \(A_{3}\) are mutually exclusive and totally exhaustive.) Now \(P\left(A / A_{1}\right)=\frac{1}{4}, P\left(A / A_{2}\right)=\frac{1}{8}\) (as given) Again it is reasonable to take the probability of answering correctly given that he knows the answer as 1 , that is, \(P\left(A / A_{3}\right)=1\) We have to find \(P\left(A_{3} / \mathrm{A}\right)\) By Bayes' theorem, we have $$ \begin{aligned} &P\left(A_{3} / A\right)=\frac{P\left(A_{3}\right) P\left(A / A_{3}\right)}{P\left(A_{1}\right) P\left(A / A_{1}\right)+P\left(A_{2}\right) P\left(A / A_{2}\right)+} \\ &\quad P\left(A_{3}\right) P\left(A / A_{3}\right) \\ &=\frac{(1 / 2) \times 1}{(1 / 3)(1 / 4)+(1 / 6)(1 / 8)+(1 / 2) \times 1}=\frac{24}{29} \end{aligned} $$

There are 3 red and 5 black balls in bag. \(A\) and 2 red and 3 black balls in bag \(B\). One ball is drawn from bag \(A\) and two from bag \(B\). Find the probability that out of the 3 balls drawn, 1 is red and 2 are black. [CBSE-96 (C)]
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