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

• A. $\dfrac{49}{4}$
• B. $\dfrac{35}{3}$
• C. $\dfrac{49}{2}$
• D. $\dfrac{35}{12}$

Hint:

A coin toss does not affect the value of $X$ here, so use only the die distribution.

Answer 1

The Correct Option is D

Step 1: See what $X$ depends on.
$X(a,b)=b$, the die face. The coin part $a$ does not affect $X$. So $X$ just behaves like a fair die showing $1$ to $6.$
Step 2: Write the distribution.
$P(X=k)=\dfrac16$ for $k=1,2,3,4,5,6.$
Step 3: Find the mean.
$E(X)=\dfrac{1+2+3+4+5+6}{6}=\dfrac{21}{6}=\dfrac72.$
Step 4: Find $E(X^2)$.
$E(X^2)=\dfrac{1+4+9+16+25+36}{6}=\dfrac{91}{6}.$
Step 5: Use the variance formula.
$\operatorname{Var}(X)=E(X^2)-[E(X)]^2=\dfrac{91}{6}-\left(\dfrac72\right)^2=\dfrac{91}{6}-\dfrac{49}{4}.$
Step 6: Combine the fractions.
Common denominator $12$: $\dfrac{182}{12}-\dfrac{147}{12}=\dfrac{35}{12}.$ \[ \boxed{\dfrac{35}{12}} \]