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.

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 $X$.

$$P(X=1)=P(X>0)-P(X>1)=1-\left(\frac{5}{6}\right)^4=1-\frac{625}{1296}=\frac{671}{1296}$$

$$P(X=2)=P(X>1)-P(X>2)=\left(\frac{5}{6}\right)^4-\left(\frac{4}{6}\right)^4=\frac{625}{1296}-\frac{256}{1296}=\frac{369}{1296}$$

$$P(X=3)=P(X>2)-P(X>3)=\left(\frac{4}{6}\right)^4-\left(\frac{3}{6}\right)^4=\frac{256}{1296}-\frac{81}{1296}=\frac{175}{1296}$$

$$P(X=4)=P(X>3)-P(X>4)=\left(\frac{3}{6}\right)^4-\left(\frac{2}{6}\right)^4=\frac{81}{1296}-\frac{16}{1296}=\frac{65}{1296}$$

$$P(X=5)=P(X>4)-P(X>5)=\left(\frac{2}{6}\right)^4-\left(\frac{1}{6}\right)^4=\frac{16}{1296}-\frac{1}{1296}=\frac{15}{1296}$$

$$P(X=6)=P(X>5)-P(X>6)=\left(\frac{1}{6}\right)^4-0=\frac{1}{1296}$$


1) is now correct after your edit based on Daniel's comment.

2) is incorrect since a) it doesn't sum to 1 and b) $X = 4, 5, 6$ doesn't sum to what you got in 1).

My approach to this question is to break it down into a bunch of simpler problems as such:

For $P(X=1)$, let's consider the $4$ dice as dice $1, 2, 3, 4$. There are $4$ possibilities, the result can have $1,2,3,$ or $4$ ones in it.

  • $1$ one: ${4 \choose 1} (\frac{1}{6})^1(\frac{5}{6})^3$
  • $2$ ones: ${4 \choose 2} (\frac{1}{6})^2(\frac{5}{6})^2$
  • $3$ ones: ${4 \choose 3} (\frac{1}{6})^3(\frac{5}{6})^1$
  • $4$ ones: ${4 \choose 4} (\frac{1}{6})^4(\frac{5}{6})^0 = (\frac{1}{6})^4$

The sum of the above will give you the $P(X=1)$.

Similarly, you can solve for the rest. Make sure that the end results sums to $1$ and $P(X=4) + P(X=5) + P(X=6) = (\frac{1}{2})^4$.

Tags:

Probability

Probability Distributions

Dice

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