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Chapter 4: Problem 17

A random variable \(X\) has the following probability distribution \(\begin{array}{rl}x: 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 \\ P(X=x) & : a & 3 a & 5 a & 7 a & 9 a & 11 a & 13 a & 15 a & 17 a\end{array}\) Then value of \(a\) is (a) \(1 / 81\) (b) \(2 / 81\) (c) \(5 / 81\) (d) \(7 / 81\)

Short Answer

Expert verified
The value of \(a\) is \(\frac{1}{81}\) (option a).

Step by step solution

01

Understanding the Probability Distribution

The given table provides the probability distribution of a discrete random variable \(X\). For a set of probabilities to be valid, the sum of all probabilities must equal 1.
02

Set Up the Equation for Total Probability

Since the sum of all probabilities \(P(X = x)\) must equal 1, we can set up the equation as follows: \[ a + 3a + 5a + 7a + 9a + 11a + 13a + 15a + 17a = 1 \] Simplifying, we find: \[ 81a = 1 \].
03

Solve for the Value of a

To find \(a\), divide both sides of the equation by 81: \( a = \frac{1}{81} \).
04

Verify the Solution

Verify by ensuring that substituting \(a = \frac{1}{81} \) into the probability distribution indeed sums to 1: \[ \frac{1}{81} + \frac{3}{81} + \frac{5}{81} + \frac{7}{81} + \frac{9}{81} + \frac{11}{81} + \frac{13}{81} + \frac{15}{81} + \frac{17}{81} = 1 \]. The sum adds up to 1, confirming our solution is correct.

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Key Concepts

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

Random Variable
A random variable is a fundamental concept in probability theory. It's essentially a variable that can take on different outcomes from a random process. Think of it like a box of possibilities linked to probabilities. In our case, we have a discrete random variable \( X \). This means its possible values are distinct and separate, like countable numbers. Each number from 0 to 8 represents a potential outcome of \( X \). Each outcome is associated with a weight or probability indicating the chance of it occurring. By understanding the random variable, you're able to predict and analyze different scenarios based on their likelihood.
Discrete Probability
Discrete probability deals specifically with outcomes that can be counted, even if the count is infinite like the set of all natural numbers. Here, the discrete probability distribution of \( X \) means every specific outcome has a probability \( P(X = x) \) attached to it. For instance, \( P(X=0) = a \), and similarly for other outcomes like 1, 2, 3, and so on, having probabilities \( 3a, 5a, 7a \), etc., respectively. This structured framework of associating probabilities with specific countable outcomes helps us manage and calculate the likelihood of different scenarios, making sense of randomness.
Sum of Probabilities
When dealing with probability distributions, an important rule is that the sum of all probabilities must equal 1. This ensures all possible outcomes are considered, and the framework is complete. In our exercise, the probabilities are set in terms of \( a \): \( a + 3a + 5a + \, ... \, + 17a \). Simplifying it, you find \( 81a \). So, for it to be a proper distribution, you set \( 81a = 1 \). It's just like ensuring your pie chart covers the whole pie鈥攅ach slice (probability) adds up to the whole (1). Solving this equation tells us exactly what \( a \) needs to be to maintain the balance: \( a = \frac{1}{81} \).
Valid Probability Distribution
A valid probability distribution must adhere to a few essential rules. First, every individual probability \( P(X = x) \) must lie between 0 and 1. This means you cannot have a negative chance or more than a certain chance of an event happening. Second, as we explored, all probabilities must sum to 1. These rules assure that the model we describe truly reflects reality. In this exercise, after finding \( a = \frac{1}{81} \), each probability \( P(X = x) \) is calculated accordingly, and they confirm with the set rule. This rigor ensures that any statistical or probabilistic work done following this distribution is grounded in mathematical truth.

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Most popular questions from this chapter

Word UNIVERSITY is arranged randomly. Then the probability that both \(I\) does not come together is: [UPSEAT-2001] |(a) \(3 / 5\) (b) \(2 / 5\) (c) \(4 / 5\) (d) \(1 / 5\) Solution (c) Total number of ways \(=\frac{10 !}{2 !}\) Favourable number of ways for \(I\) come together is \(9 !\) Thus, probability that \(I\) come together $$ =\frac{9 ! \times 2 !}{10 !}=\frac{2}{10}=\frac{1}{5} $$

Urn \(A\) contains 6 red and 4 black balls and urn \(B\) contains 4 red and 6 black balls. One ball is drawn at random from urn \(A\) and placed in urn \(B\). Then 1 ball is drawn at random from urn \(B\) and placed in urn \(A\). If 1 ball is now drawn at random from urn \(A\), the probability that it is found to be red is \([\) IIT-1988] (a) \(\frac{32}{55}\) (b) \(\frac{21}{55}\) (c) \(\frac{19}{55}\) (d) None of these (a) Let the events are \(R_{1}\) : 'a red ball is drawn from urn \(A\) and placed in \(B\) ' \(B_{1}:\) 'a black ball is drawn from urn \(A\) and placed in \(B\) ' \(R_{2}:\) 'a red ball is drawn from urn \(B\) and placed in \(A^{\prime}\) \(B_{2}:\) 'a black ball is drawn from urn \(B\) and placed in \(A\) 'Urn \(A\) contains 6 red and 4 black balls and urn \(B\) contains 4 red and 6 black balls. One ball is drawn at random from urn \(A\) and placed in urn \(B\). Then 1 ball is drawn at random from urn \(B\) and placed in urn \(A\). If 1 ball is now drawn at random from urn \(A\), the probability that it is found to be red is \([\) IIT-1988] (a) \(\frac{32}{55}\) (b) \(\frac{21}{55}\) (c) \(\frac{19}{55}\) (d) None of these (a) Let the events are \(R_{1}\) : 'a red ball is drawn from urn \(A\) and placed in \(B\) ' \(B_{1}:\) 'a black ball is drawn from urn \(A\) and placed in \(B\) ' \(R_{2}:\) 'a red ball is drawn from urn \(B\) and placed in \(A^{\prime}\) \(B_{2}:\) 'a black ball is drawn from urn \(B\) and placed in \(A\) '

If the letters of the word ATTEMPT are written down at random. Find the probability if (i) all \(T\) s are together (ii) no two \(T\) s are together (a) (i) \(1 / 7\) (ii) \(2 / 7\) (b) (i) \(2 / 7\) (ii) \(1 / 7\) (c) (i) \(2 / 7\) (ii) \(3 / 7\) (d) (i) \(3 / 7\) (ii) \(1 / 7\)

A letter is known to have come from either TATANAGAR or CALCUTTA. On the envelope, just two consecutive letters. TA are visible. The probability that the letters have come from CALCUTTA is (a) \(1 / 3\) (b) \(4 / 11\) (c) \(5 / 12\) (d) None of these (b) Let \(A\) : 'the event that letters came from TATANAGAR' B: 'the event that letters came from CALCUTTA' \(C\) : 'the event that two consecutive letters visible by \(T A^{\prime}\) Then \(P(A)=1 / 2, P(B)=1 / 2, P(C / B)=1 / 7\) Hence, by Bayes' theorem $$ P(B / C)=\frac{P(B) \times P(C / B)}{P(A) P(C / A)+P(B) P(C / B)}=\frac{4}{11} $$

A coin is tossed twice. Find the probability distribution of the number of heads.
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