Problem 20 - Algebra 2 Course in Mathematics for the IIT-JEE and Other Engineering Exams

Question

For a binomial variate \(X\) if \(n=5\) and \(P(X=1)=\) \(8 P(\mathrm{X}=3)\), then \(P\) is (a) \(4 / 5\) (b) \(1 / 5\) (c) \(1 / 3\) (d) \(2 / 3\) tion $$ \begin{aligned} &\text { (b) }{\underline{\phantom{xx}}}^{5} C_{1} q^{4} p^{1}=8 \times{ }^{5} C_{3} q^{2} p^{3} \\ &\Rightarrow q=4 P \\ &\Rightarrow 1-p=4 P \\ &\Rightarrow P=1 / 5 \end{aligned} $$

Step-by-Step Solution

Answer

The probability \( P \) is \( \frac{1}{5} \), which corresponds to option (b).

Step 1: Understand the Binomial Probability Formula

For a binomial distribution, the probability of getting exactly k successes in n trials is given by the formula \( P(X=k) = \binom{n}{k} p^k (1-p)^{n-k} \), where \( p \) is the probability of success on an individual trial.

Step 2: Set the Problem Equation

We are given that \( P(X=1) = 8 \times P(X=3) \). Using the binomial formula, we write two separate equations: \( P(X=1) = \binom{5}{1} p^1 (1-p)^4 \) and \( P(X=3) = \binom{5}{3} p^3 (1-p)^2 \).

Step 3: Substitute the Binomial Coefficients

Calculate and substitute the coefficients: \( \binom{5}{1} = 5 \) and \( \binom{5}{3} = 10 \). This simplifies our probabilities to \( P(X=1) = 5p(1-p)^4 \) and \( P(X=3) = 10p^3(1-p)^2 \).

Step 4: Formulate the Equation

Substitute into the given relationship: \( 5p(1-p)^4 = 8 \times 10p^3(1-p)^2 \). Simplify this to: \( 5(1-p)^4 = 80p^2(1-p)^2 \).

Step 5: Solve for p

Divide both sides by \((1-p)^2\) (assuming \( p<1\) so \(1-p\) is nonzero), leading to \( 5(1-p)^2 = 80p^2 \). Rewrite as \( 5(1-2p+p^2) = 80p^2 \) or \( 5 - 10p + 5p^2 = 80p^2 \).

Step 6: Simplify the Equation

Rearrange the equation to form a quadratic: \( 75p^2 + 10p - 5 = 0 \). Divide the entire equation by 5 to simplify it: \( 15p^2 + 2p - 1 = 0 \).

Step 7: Solve the Quadratic

Use the quadratic formula \( p = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 15, b = 2, \) and \( c = -1 \). Calculate the discriminant: \( b^2 - 4ac = 4 + 60 = 64 \). Thus, the solutions for \( p \) are \( p = \frac{-2 \pm 8}{30} \).

Step 8: Evaluate Possible Solutions

The possible solutions are \( p = \frac{6}{30} = \frac{1}{5} \) or \( p = \frac{-10}{30} = -\frac{1}{3} \). Since probability cannot be negative, only \( p = \frac{1}{5} \) is valid.

Key Concepts

Binomial Probability Formula

In probability theory, the binomial distribution is a widely used probability distribution. It describes the number of successes in a fixed number of independent and identical trials, each with the same probability of success. When we're dealing with binomial distributions, it's essential to grasp the binomial probability formula, which calculates the likelihood of obtaining a specific number of successes.

This formula is expressed as:\[ P(X=k) = \binom{n}{k} p^k (1-p)^{n-k} \]where:

• \( P(X=k) \): Probability of getting exactly \( k \) successes
• \( \binom{n}{k} \): Binomial coefficient, representing the number of ways to choose \( k \) successes from \( n \) trials
• \( p \): Probability of success on an individual trial
• \( (1-p) \): Probability of failure on an individual trial
• \( n \): Total number of trials

Using the binomial formula, you can determine outcomes that seem challenging by breaking them into manageable parts. Calculating each component becomes easy once you grasp how each part contributes to formulating the probability.

Quadratic Equation

In the process of working with probability problems, we sometimes encounter quadratic equations. These arise when probabilities combine in ways that form a squared component. For example, in our exercise, after simplifying using the binomial formula to express probabilities, we reached a quadratic equation.

A quadratic equation generally takes the form:\[ ax^2 + bx + c = 0 \]where:

• \( a \), \( b \), and \( c \) are constants
• \( x \) represents the variable or unknown we're solving for

Rewriting probabilities sometimes results in equations with terms that include squares, leading to the need to solve quadratic equations.

In our example, manipulating terms from the binomial distribution led us to solve the expression:\[ 15p^2 + 2p - 1 = 0 \]This form allowed us to apply the quadratic formula to find the probability value \( p \) accurately.

Solving Equations

In probability exercises, solving equations is crucial to finding specific outcomes or probabilities. Once equations are set up, the next step involves determining the value of unknowns that satisfy the conditions laid out.

A common method involves the quadratic formula, which provides solutions to quadratic equations:\[ p = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]This formula helps us find the roots or solutions of a quadratic equation involving any polynomial expression set equal to zero.

By computing the discriminant, \( b^2 - 4ac \), and substituting into the quadratic formula, we can evaluate the validity of possible solutions. This approach resolves the unknowns comprehensively, reducing complexities into straightforward calculations.

In practical scenarios, such operations demand attentiveness to ensure positive probabilities, underscoring thoughtful equation formulation and solution testing.

Probability of Success

In probability theory, especially involving binomial distributions, the 'probability of success' refers to the likelihood that an individual trial will result in a successful outcome. Understanding this concept is crucial in calculating binomial probabilities.

The probability of success is denoted by \( p \), whereas the probability of failure in a trial is expressed as \( 1-p \). In scenarios involving multiple trials, maintaining consistent success probabilities across each trial is paramount.

• Defining Success: Clearly determine what counts as a success. It's the foundation for calculating corresponding probabilities.
• Consistency: Ensure that the conditions under which trials are conducted remain identical throughout the process. This consistency upholds the validity of the binomial model.

In solving our exercise, the given condition \( P(X=1) = 8 \times P(X=3) \) led us to manipulate probabilities and equations to evaluate \( p \). Concluding with a valid probability, such as \( p = \frac{1}{5} \), confirms that the mathematical framework correctly interpreted the probability scenario.