Question

We roll a fair 8-sided die five times. (A fair 8-sided die is equally likely to be 1, 2, 3, 4, 5, 6, 7, or 8.) (a) What is the probability that at least one of the rolls is a 3? (b) Let X be the number of different values rolled. For example, if the five rolls are 2, 3, 8, 8, 7, then X = 4 (since four different values were rolled: 2,3,7,8). Find E[X].

Answer 1

a) Probability that at least one of the rolls is a 3= 1 - Probability that none of the rolls is a 3

Probability of getting a 3= 1/8

Probability of not getting a 3= 1-1/8 = 7/8

Since each roll is independent therefore probability of not getting a 3 in any 5 rolls=

5

Therefore probability of getting at least one of the rolls as 3=

$1- \left ( \frac{7}{8} \right )^{5}$

b) The values of X can be from the set X = {1,2,3,4,5}

Total sample space size=$8^5$

P(X = 1) = ic 85

$P(X=2)=\frac{ \ _{2}^{8}\textrm{C} \ 2^3}{8^5}$

$P(X=3)=\frac{ \ _{3}^{8}\textrm{C} \ 3^2}{8^5}$

$P(X=4)=\frac{ \ _{4}^{8}\textrm{C} \ 4}{8^5}$

$P(X=5)=\frac{ \ _{5}^{8}\textrm{C} }{8^5}$

Now,

$E(X)=1\frac{ \ _{1}^{8}\textrm{C} }{8^5}+2 \frac{ \ _{2}^{8}\textrm{C} \ 2^3 }{8^5} + 3\frac{ \ _{3}^{8}\textrm{C} \ 3^2 }{8^5} + 4\frac{ \ _{4}^{8}\textrm{C} \ 4}{8^5} + 5\frac{ \ _{5}^{8}\textrm{C} }{8^5}$

$E(x)=\frac{3368}{32768}=0.1027$