Question

Jessica has a four-sided die, a six-sided die, and a 12-sided die. She rolls the three dice once. Let Z be the number of fours showing. Find E[Z] using the indicator method.

Answer 1

Answer:

The value of E [Z]  is 0.50.

Step-by-step explanation:

The expected value is computed using the formula:

$E(X)=\sum x\cdot P(X= x)$

Denote the three dices as follows:

X₁ = a four-sided die

X₂ = a six-sided die

X₃ = a 12-sided die

Define the indicator variables as follows:

$I_{1}=\left \{ {{1;\ X_{1}=4} \atop {0;\ X_{1}\neq 4}} \right. \\\\I_{2}=\left \{ {{1;\ X_{2}=4} \atop {0;\ X_{2}\neq 4}} \right. \\\\I_{3}=\left \{ {{1;\ X_{3}=4} \atop {0;\ X_{3}\neq 4}} \right.$

The probability of rolling a 4 in the three dices are as follows:

$P(X_{1}=4)=\frac{1}{4}\\\\P(X_{2}=4)=\frac{1}{6}\\\\P(X_{3}=4)=\frac{1}{12}$

Compute the value of E [Z] as follows:

$E(Z)=E(I_{1})+E(I_{2})+E(I_{3})$

         $=(1\times\frac{1}{4})+(1\times\frac{1}{6})+(1\times\frac{1}{12})\\\\=\frac{3+2+1}{12}\\\\=\frac{6}{12}\\\\=\frac{1}{2}\\\\=0.50$

Thus, the value of E [Z]  is 0.50.