PMF of throwing a die 4 times We throw a fair 6-sided die independently four times and let $X$ denote the minimal value rolled.


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*What is the probability that $X \ge 4$?

*Compute the PMF of $X$.

*Determine the mean and variance of $X$.
My attempt:


*

*$(1/2)^4$ because that means each of the $4$ rolls, you either get a $4, 5$ or $6$.

*for $X=1: (1/6)(6/6)(6/6)(6/6)$
for $X=2: (1/6)(5/6)(5/6)(5/6)$
for $X=3: (1/6)(4/6)(4/6)(4/6)$
for $X=4: (1/6)(3/6)(3/6)(3/6)$
for $X=5: (1/6)(2/6)(2/6)(2/6)$
for $X=6: (1/6)(1/6)(1/6)(1/6)$

*I can calculate this once I know I did the PMF correctly
Did I do (1) and (2) correctly?

We throw a fair 6-sided die independently four times and let $X$ denote the minimal value rolled.

What is the probability that $X \ge 4$?

For each roll, the value must be greater than three. This event has probability $\frac{3}{6}=\frac{1}{2}$. As this must occur on each of $4$ rolls, we have:
$$P(X\ge4)=\left(\frac{1}{2}\right)^4=\frac{1}{16}$$

Compute the PMF of $X$.

$$P(X=1)=P(X>0)-P(X>1)=1-\left(\frac{5}{6}\right)^4=1-\frac{625}{1296}=\frac{671}{1296}$$
$$P(X=2)=P(X>1)-P(X>2)=\left(\frac{5}{6}\right)^4-\left(\frac{4}{6}\right)^4=\frac{625}{1296}-\frac{256}{1296}=\frac{369}{1296}$$
$$P(X=3)=P(X>2)-P(X>3)=\left(\frac{4}{6}\right)^4-\left(\frac{3}{6}\right)^4=\frac{256}{1296}-\frac{81}{1296}=\frac{175}{1296}$$
$$P(X=4)=P(X>3)-P(X>4)=\left(\frac{3}{6}\right)^4-\left(\frac{2}{6}\right)^4=\frac{81}{1296}-\frac{16}{1296}=\frac{65}{1296}$$
$$P(X=5)=P(X>4)-P(X>5)=\left(\frac{2}{6}\right)^4-\left(\frac{1}{6}\right)^4=\frac{16}{1296}-\frac{1}{1296}=\frac{15}{1296}$$
$$P(X=6)=P(X>5)-P(X>6)=\left(\frac{1}{6}\right)^4-0=\frac{1}{1296}$$
