What is the probability $\{1,2,3,4,5\}$ occurs exactly once when we throw a die $5$ times?

Question

If you throw a fair die 5 times, then
a. What is probability  {1,2,3,4,5} occur exactly once ?
b. What is probability least one of {1,2,3,4,5} occur?
c. What is expected value least one of {1,2,3,4,5} occur?
d. What is the variance of least one of {1,2,3,4,5}?

My solution:
a. P(X = 5) = 5C5 (5/6)^4 (1/6)^1 = 627/7776= 0.0803
b. P(X = 0) = (1/6)^5
P (at least one of number <= 5) = 1-(1/6)^5= 0.9999
c. E(X) = np = 5*(0.9999)
d. Var(X) = np(1-p)= 5*(0.9999)(1-0.9999)

I am not sure I get right or wrong, I couldn't find solution to check my work,I am really appreciate your helps. Thanks.

Answer 1

If you throw a fair dice 5 times, then a. What is probability that we will see exactly one time of a number <= 5? b. What is probability that we will see exactly at least one of a number <= 5? c. What is expected number of seeing a number <= 5? d. What is the variance of seeing a number <= 5?

My solution:

a. P(X = 5) = 5C5 (5/6)^4 (1/6)^1 = 627/7776= 0.0803

This is almost the probability that you obtain an arrangement of one of {1,2,3,4,5} and four other number however it over counts common cases.   You need to apply the principle of inclusion and exclusion.

b. P(X = 0) = (1/6)^5 P (at least one of number <= 5) = 1-(1/6)^5= 0.9999

Yes; with rounding. $\tfrac{7775}{7776}\approx. 0.9998{\small 7\tiny 14\ldots} $

c. E(X) = np = 5*(0.9999)

No; not quite.   Let $X_n$ be the indicator that you see $n$ occur. $\mathsf P(X_n=1) =1-(\tfrac 5 6)^5$.   Then $$\mathsf E(X_n) ~=~ 0+1\times\dfrac{6^5-5^5}{6^5}$$

So since the count of such occurrences is equal to the sum of the indicators, we use the Linearity of Expectation:

$$\begin{align}\mathsf E(X) =&~ \sum_{n=1}^5 \mathsf E(X_n) \\ =&~ \dfrac {5~( 6^5-5^5)}{6^5}\end{align}$$

d. Var(X) = np(1-p)= 5*(0.9999)(1-0.9999)

It is a little more complicated than that.   This is not a binomial distribution.

Now because when $n\neq m$ we have $\mathsf E(X_nX_m)~=~\mathsf P(X_n=1,X_m=1) ~=~ 1-\tfrac {4^5}{6^5}$, so then:

$$\begin{align}\mathsf {Var}(X) =&~ \mathsf E[(\sum_{n=1}^5X_n)^2] - \mathsf E[\sum_{n=1}^5 X_n]^2 \\[1ex] =&~ \left(\sum_{n=1}^5 \mathsf E(X_n^2) + \raise{1ex}\mathop{\sum_{n=1}^5\sum_{m=1}^5}_{n\neq m}\mathsf E(X_nX_m)\right) - \mathsf E(X)^2 \end{align}$$