Question

The discrete random variables \(X\) and \(Y\) are independent from one another and are defined as
\[ X\sim B(16,0.25) \] and
\[ Y\sim P(2). \] Then the sum of the variances of \(X\) and \(Y\) is

• A. \(4\)
• B. \(5\)
• C. \(6\)
• D. \(2\)

Hint:

Remember the standard variance formulas:
For \(B(n,p)\): \(\mathrm{Var}=npq\) and for \(P(\lambda)\): \(\mathrm{Var}=\lambda\).

Answer 1

The Correct Option is B

Step 1: Find the variance of \(X\).
Given,
\[ X\sim B(16,0.25) \] which is a binomial distribution with
\[ n=16,\qquad p=0.25 \]
For a binomial distribution, variance is given by
\[ \mathrm{Var}(X)=npq \] where
\[ q=1-p \]
Thus,
\[ q=1-0.25=0.75 \]
Therefore,
\[ \mathrm{Var}(X)=16\times0.25\times0.75 \]
\[ =4\times0.75 \]
\[ =3 \]

Step 2: Find the variance of \(Y\).
Given,
\[ Y\sim P(2) \] which is a Poisson distribution with parameter \(\lambda=2\).
For a Poisson distribution,
\[ \mathrm{Var}(Y)=\lambda \]
Hence,
\[ \mathrm{Var}(Y)=2 \]

Step 3: Find the sum of variances.
\[ \mathrm{Var}(X)+\mathrm{Var}(Y) \]
\[ =3+2 \]
\[ =5 \]

Step 4: Final conclusion.
Hence, the required sum is
\[ \boxed{5} \]