Question

The discrete random variables \(X\) and \(Y\) are independent from one another and are defined as \(X \sim B(n_1, 0.5)\) and \(Y \sim B(n_2, 0.4)\). If the variance of both \(X\) and \(Y\) is 6, then \(\sqrt{n_1+n_2} =\)

• A. 7
• B. 6
• C. 5
• D. 4

Hint:

For independent binomial random variables, use \(\text{Var}(X) = n p (1-p)\) to find unknown \(n\) or \(p\), then combine as needed.

Answer 1

The Correct Option is A

Step 1: Write down the variance condition for each binomial variable.
\[ \text{Var}(X) = n_1 p_1(1-p_1), \qquad \text{Var}(Y) = n_2 p_2(1-p_2) \]

Step 2: Plug in the two probabilities and set each variance to 6.
For \(X\), \(p_1=0.5\) gives \(p_1(1-p_1)=0.25\), so \(n_1 = \frac{6}{0.25}=24\).
For \(Y\), \(p_2=0.4\) gives \(p_2(1-p_2)=0.24\), so \(n_2 = \frac{6}{0.24}=25\).

Step 3: Add the two values of n.
\[ n_1+n_2 = 24+25 = 49 \]

Step 4: Take the square root.
\[ \sqrt{n_1+n_2} = \sqrt{49} = 7 \]
\[ \boxed{7} \]