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

Question

The value of \(C\) for which \(P(X=k)=C k^{2}\) can serve as the probability function of a random variable \(X\) that takes \(0,1,2,3,4\) is [EAMCET-1994] (a) \(1 / 30\) (b) \(1 / 10\) (c) \(1 / 3\) (d) \(1 / 15\) (a) \(\sum_{k=0}^{4} P(X=k)=1 \Rightarrow \sum_{k=0}^{4} C_{k}^{2}=1\) \(\Rightarrow C\left(1^{2}+2^{2}+3^{2}+4^{2}\right)=1\) \(\Rightarrow C=\frac{1}{20}\)Solution (b) Here mean \(=n p\) and variance \(=n p q\) $$ \begin{gathered} \therefore \frac{P(X=k)}{P(X=k-1)}=\frac{{ }^{n} C_{k}(p)^{k}(q)^{n-k}}{{ }^{n} C_{k-1}(p)^{k-1}(q)^{n-k-1}}=\frac{{ }^{n} C_{k}}{{ }^{n} C_{k-1}} \times \frac{p}{q} \\ \quad \therefore \frac{P(X=k)}{P(X=k-1)}=\frac{n-k+1}{k} \times \frac{p}{q} \end{gathered} $$ 7\. Two cards are drawn successively with replacement from a well-shuffled deck of 52 cards then the mean of the number of aces is (a) \(1 / 13\) (b) \(3 / 13\) (c) \(2 / 13\) (d) None of these Solution (c) Let \(X\) denote a random variable which is the number of aces. Clearly, \(X\) takes values 1,2 . $$ \begin{aligned} &\therefore p=\frac{4}{52}=\frac{1}{13}, q=1-\frac{1}{13}=\frac{12}{13} \\ &P(X=1)=2 \times\left(\frac{1}{13}\right)\left(\frac{12}{13}\right)=\frac{24}{169} \\ &P(X=2)=2 \times\left(\frac{1}{13}\right)^{2}\left(\frac{12}{13}\right)^{0}=\frac{1}{169} \\\ &\text { Mean }=\sum P_{i} X_{i}=\frac{24}{169}+\frac{2}{169}=\frac{26}{169}=\frac{2}{13} \end{aligned} $$

Step-by-Step Solution

Answer

The value of \( C \) is \( \frac{1}{30} \) (option a).

Step 1: Understand the Problem

We need to find the value of \( C \) such that the distribution \( P(X=k)=Ck^2 \) over the values \( 0, 1, 2, 3, 4 \) is a valid probability distribution.

Step 2: Set Up the Equation

For a distribution to be valid, the sum of probabilities must equal 1. Therefore, we need \( \sum_{k=0}^{4} P(X=k) = 1 \). Substitute the given \( P(X=k) = Ck^2 \) into the sum: \( C (0^2 + 1^2 + 2^2 + 3^2 + 4^2) = 1 \).

Step 3: Calculate the Sum of Squares

Calculate \( 0^2, 1^2, 2^2, 3^2, 4^2 \), which are \( 0, 1, 4, 9, \) and \( 16 \) respectively. Add these values: \( 0 + 1 + 4 + 9 + 16 = 30 \).

Step 4: Solve for C

Substitute the sum of squares back into the equation: \( C \times 30 = 1 \). Solve for \( C \) by dividing both sides by 30: \( C = \frac{1}{30} \).

Step 5: Verify the Solution

Check if the computed \( C \) correctly establishes a probability distribution. Since \( \sum_{k=0}^{4} Ck^2 = 1 \) holds true with \( C = \frac{1}{30} \), our solution is verified.

Key Concepts

Random Variable

A random variable is a fundamental concept in probability and statistics. Think of it as a variable that can take on different numerical values, each associated with a certain probability. Imagine flipping a coin, the outcome (heads or tails) is a random variable, because each possibility occurs randomly.
Here's an easy way to understand it:

• Random variables are often denoted by letters like X or Y.
• They can be discrete (taking specific values like 0, 1, 2...) or continuous (any value in a range like 1.5, 2.8...).
• Every outcome has a probability linked to it. For example, in a dice roll, the random variable representing the number rolled can be 1, 2, ..., 6, each with a probability of 1/6.

In probability distributions, a random variable is linked with a method that outlines how probabilities are distributed among all potential outcomes. Understanding random variables helps us quantify uncertainty in experiments or data models.

Binomial Distribution

Binomial distribution is a particular type of discrete probability distribution that models the number of successes in a series of independent and identical trials. It's akin to repeatedly flipping a coin a set number of times and counting the number of heads you get, where each flip is a trial.
Here are the key features:

• It has two parameters: the number of trials (n) and the probability of success in a single trial (p).
• Each trial must be independent, meaning the outcome of one trial doesn't affect another.
• The outcomes of a binomial distribution are discrete, like flipping a coin where outcomes could be 0 heads, 1 head, etc.

This distribution is used in various fields, such as quality control, to predict pass/fail outcomes. The probability of getting a certain number of successes (like rolling 3 heads in 5 coin tosses) can be determined using the formula: \[P(X=k) = \binom{n}{k} p^k (1-p)^{n-k}\]where \(\binom{n}{k}\) is the binomial coefficient and represents the number of ways to choose \(k\) successes in \(n\) trials. Understanding binomial distributions can greatly enhance your ability to work with data involving binary outcomes.

Mean of Random Variable

The mean of a random variable, often called the expected value, gives the average outcome if you were to repeat the random experiment many times. It tells us the long-run average value of the random variable.
Here's how you can interpret the mean:

• For discrete random variables, the mean is calculated by summing up the products of each value of the variable and its probability. Mathematically, this is expressed as \(E(X) = \sum x_i P(x_i)\), where \(x_i\) are the values and \(P(x_i)\) their probabilities.
• In a binomial distribution, the mean can be found using the formula \(E(X) = np\), where \(n\) is the number of trials and \(p\) the probability of success in each trial.
• The mean provides insight into the "center" of the distribution, offering a prediction of outcomes over numerous trials.

For instance, in the card example mentioned where two cards are drawn, the mean helps to predict the average number of aces drawn. Understanding the mean helps in making data-driven decisions, as you can predict expected outcomes.