Question:
Consider the random experiment of throwing a die and tossing a coin. Let
\[
\{(a,b)\mid a\in\{H,T\},\ b\in\{1,2,\ldots,6\}\}
\]
denote the outcomes of the experiment. If $X$ is a random variable defined by
\[
X(a,b)=b,
\]
then the variance of $X$ is
Consider the random experiment of throwing a die and tossing a coin. Let
\[
\{(a,b)\mid a\in\{H,T\},\ b\in\{1,2,\ldots,6\}\}
\]
denote the outcomes of the experiment. If $X$ is a random variable defined by
\[
X(a,b)=b,
\]
then the variance of $X$ is
Show Hint
A coin toss does not affect the value of $X$ here, so use only the die distribution.
Updated On: Jun 3, 2026
- $\dfrac{49}{4}$
- $\dfrac{35}{3}$
- $\dfrac{49}{2}$
- $\dfrac{35}{12}$
Show Solution
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The Correct Option is D
Solution and Explanation
Step 1: Concept
The random variable depends only on the die outcome, so it has the same distribution as a fair die.
Step 2: Meaning
Thus \[ P(X=k)=\frac16,\qquad k=1,2,\ldots,6. \]
Step 3: Analysis
Compute \[ E(X)=\frac{1+2+3+4+5+6}{6} =\frac72. \] Also, \[ E(X^2) = \frac{1^2+2^2+3^2+4^2+5^2+6^2}{6} = \frac{91}{6}. \] Hence \[ \operatorname{Var}(X) = E(X^2)-[E(X)]^2 = \frac{91}{6}-\left(\frac72\right)^2. \] \[ = \frac{182-147}{12} = \frac{35}{12}. \]
Step 4: Conclusion
Therefore, \[ \operatorname{Var}(X)=\frac{35}{12}. \]
Final Answer: (D)
The random variable depends only on the die outcome, so it has the same distribution as a fair die.
Step 2: Meaning
Thus \[ P(X=k)=\frac16,\qquad k=1,2,\ldots,6. \]
Step 3: Analysis
Compute \[ E(X)=\frac{1+2+3+4+5+6}{6} =\frac72. \] Also, \[ E(X^2) = \frac{1^2+2^2+3^2+4^2+5^2+6^2}{6} = \frac{91}{6}. \] Hence \[ \operatorname{Var}(X) = E(X^2)-[E(X)]^2 = \frac{91}{6}-\left(\frac72\right)^2. \] \[ = \frac{182-147}{12} = \frac{35}{12}. \]
Step 4: Conclusion
Therefore, \[ \operatorname{Var}(X)=\frac{35}{12}. \]
Final Answer: (D)
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