My class and I are confused about an exercises our teacher gave us:
A teacher fills in the report sheets in the following way. For each student he rolls three dice and determines the grade as the greatest of the occuring numbers.
a) What is the probabilty of a 6? b) What are the probabilities of the grades 1 to 5?
Thank you very much for your answers.
Kind regards
On
Let $X$ be the random variable in question (the maximum of the die values). It is easiest to compute $\psi_i=Prob(X≤i)$. To get $X≤i$ we need each die to show one of $\{1,\cdots, i\}$ hence $$\psi_i=\frac {(i)^3}{6^3}$$
The probability you seek, $P_i=Prob(X=i)$ is given by $P_i=\psi_{i}-\psi_{i-1}$. Thus $$P_i= \frac {(i)^3}{6^3}-\frac {(i-1)^3}{6^3}=\frac {i^3-(i-1)^3}{6^3}$$ Numerically, we have $$\{P_i\}_{i=1}^6=\{0.00462963,0.032407407,0.087962963,0.171296296,0.282407407,0.421296296\}$$
On
getting 6 means that it wasn't the case that, no roll better was than a 5.
$P(6) = 1 - \left(\frac {5}{6}\right)^3$
Extending this logic, getting means not the case that all three dice were 4 or less and not a 6
$P(5) = 1 - \left(\frac {4}{6}\right)^3 - P(6) = \left(\frac {5}{6}\right)^3 - \left(\frac {4}{6}\right)^3$
$P(n) = \left(\frac {n}{6}\right)^3 - \left(\frac {n-1}{6}\right)^3$
As I see, the probability to get a 6 for a student is $1-(\frac{5}{6})^3$. And get 1 to 5 is $(\frac{5}{6})^3$. What is your question?