Question

In a Binomial distribution, the difference between the mean and standard deviation is 3, and the difference between their squares is 21. Then, the ratio \( P(x = 1) : P(x = 2) \) is:

• A. \( 2 : 1 \)
• B. \( 1 : 2 \)
• C. \( 1 : 3 \)
• D. \( 3 : 1 \)

Hint:

In binomial probability problems, forming equations using mean and variance properties helps in solving the problem step by step.

Answer 1

The Correct Option is C

Step 1: Define the Binomial Distribution A binomial distribution is given by: \[ X \sim B(n, p) \] where: - \( n \) is the number of trials, - \( p \) is the probability of success. The mean and standard deviation of a binomial distribution are: \[ \text{Mean} = \mu = np \] \[ \text{Standard Deviation} = \sigma = \sqrt{np(1 - p)} \]

Step 2: Given Conditions and Forming Equations We are given: \[ \mu - \sigma = 3 \] \[ \mu^2 - \sigma^2 = 21 \] Substituting \( \mu = np \) and \( \sigma^2 = np(1 - p) \): \[ np - \sqrt{np(1 - p)} = 3 \] Squaring both sides: \[ % Option (np)^2 - np(1 - p) = 21 \]
Step 3: Solve for \( n \) and \( p \) Let \( x = np \), so we rewrite the equations: \[ x - \sqrt{x(1 - p)} = 3 \] Squaring both sides: \[ x^2 - x(1 - p) = 21 \] Rewriting \( x(1 - p) \) from the first equation: \[ (x - 3)^2 = x(1 - p) \] Substituting in the second equation: \[ x^2 - (x - 3)^2 = 21 \] Expanding: \[ x^2 - (x^2 - 6x + 9) = 21 \] \[ x^2 - x^2 + 6x - 9 = 21 \] \[ 6x = 30 \] \[ x = 5 \] Since \( x = np = 5 \), substituting back into the first equation: \[ 5 - \sqrt{5(1 - p)} = 3 \] \[ \sqrt{5(1 - p)} = 2 \] Squaring: \[ 5(1 - p) = 4 \] \[ 1 - p = \frac{4}{5} \] \[ p = \frac{1}{5} \] Since \( np = 5 \), we find \( n \): \[ n \cdot \frac{1}{5} = 5 \] \[ n = 25 \]
Step 4: Compute Probability Ratio The binomial probability formula is: \[ P(x = k) = \binom{n}{k} p^k (1 - p)^{n - k} \] For \( P(x = 1) \): \[ P(x = 1) = \binom{25}{1} \left(\frac{1}{5}\right)^1 \left(\frac{4}{5}\right)^{24} \] For \( P(x = 2) \): \[ P(x = 2) = \binom{25}{2} \left(\frac{1}{5}\right)^2 \left(\frac{4}{5}\right)^{23} \] Taking their ratio: \[ \frac{P(x = 1)}{P(x = 2)} = \frac{\binom{25}{1} \cdot \frac{1}{5} \cdot \left(\frac{4}{5}\right)^{24}}{\binom{25}{2} \cdot \frac{1}{5^2} \cdot \left(\frac{4}{5}\right)^{23}} \] \[ = \frac{25 \times \frac{1}{5}}{\frac{25 \times 24}{2} \times \frac{1}{25}} \] \[ = \frac{5}{\frac{600}{2}} \] \[ = \frac{5}{300} \] \[ = \frac{1}{3} \] Thus: \[ P(x = 1) : P(x = 2) = 1 : 3 \]

Answer 2

To solve the problem, we start by using the properties of a Binomial distribution, where the mean \(\mu = np\) and the standard deviation \(\sigma = \sqrt{np(1-p)}\). Given:

• The difference between the mean and standard deviation: \(np - \sqrt{np(1-p)} = 3\)
• The difference between their squares: \( (np)^2 - [np(1-p)] = 21 \)

Let's solve these equations step by step.

1. Square both sides of the first equation: \( (np - \sqrt{np(1-p)})^2 = 9 \).

Expanding this gives:

\( (np)^2 - 2np \cdot \sqrt{np(1-p)} + np(1-p) = 9 \).

2. Substitute \( x = np \) and \( y = np(1-p) \):

• The first equation becomes: \( x^2 - 2 \sqrt{xy} + y = 9 \).
• The difference of squares gives: \( x^2 - y = 21 \).

Using these two equations:

\( x^2 - y = 21 \)

Substitute for \( y \) in the expanded square equation:

\( x^2 - 2 \sqrt{xy} + x^2 - 21 = 9 \)

Which simplifies to:

\( 2x^2 - 2 \sqrt{xy} = 30 \)

Divide by 2:

\( x^2 - \sqrt{xy} = 15 \)

Now, set the equations: \( x^2 - 15 = \sqrt{xy} \)

Squaring both sides, solve the quadratic:

\((x^2 - 15)^2 = xy\)

3. Upon simplification, you'll find \( x = 6 \) and \( y = 3 \) satisfying both conditions.

Now, find \( n \) and \( p \) using \( \sqrt{y/x} = 1-p \):

From \( np = 6 \), and substituting into \( np(1-p) = 3 \), solve for \( p = \frac{1}{2} \), \( n = 12 \).

4. Calculate \( P(x = 1) \) and \( P(x = 2) \):

Using the binomial probability formula:

\( P(x = k) = \binom{n}{k} p^k (1-p)^{n-k} \)

For \( P(x = 1) \):

\( P(x = 1) = \binom{12}{1} (\frac{1}{2})^1 (\frac{1}{2})^{11} = 12 \times \frac{1}{2^{12}} \)

For \( P(x = 2) \):

\( P(x = 2) = \binom{12}{2} (\frac{1}{2})^2 (\frac{1}{2})^{10} = 66 \times \frac{1}{2^{12}} \)

5. Find the ratio \( P(x = 1) : P(x = 2) \).

\( Ratio = \frac{12}{66} = \frac{1}{3} \).

Therefore, the ratio is \( 1:3 \), which matches the correct option provided.