Question

Problem 1. Suppose that you roll an 8-sided die until you get an 8. Let G denote the number of rolls that this takes. (a) Write down the probability mass function of G (b) Write a closed-form expression for P(G 2 6) (i.e. do not just write it as an infinite sum) (c) Do calculations to show that P(G2 10 | G> 6) P(G2 4).

Answer 1

The probability of getting an 8 in any roll is $\dfrac{1}{8}$.

a) G follows a geometric distribution with parameter $\dfrac{1}{8}$.

P(G = g)
means the probability of getting an 8 on the gth roll and not getting an 8 in each of the throws before it. Since each roll is independent this probability is :
1-8

Thus, $P(G=g)=\dfrac{1}{8} \cdot \left(\dfrac{7}{8} \right )^{g-1}$ is the pmf of G. Also

G>1

b)

P(G> 6) 719-1 9-6

                                                                                                       

Ri 7-8 7-8 1-8 1-8

c) $P(G \geq 10 |G>6)=\dfrac{P(G\geq 10 \ \text{and} \ \ G>6)}{P(G>6)}=\dfrac{P(G \geq 10)}{P(G \geq 7)}$

Let's calculate all the required values:

9-4 719-1 P(G > 4) に4

                                                                                                       

Ri 7-8 7-8 1-8 1-8

$P(G \geq 7)=\sum_{g=7}^{\infty}P(G=g)=\sum_{g=7}^{\infty}\dfrac{1}{8} \cdot \left(\dfrac{7}{8} \right )^{g-1}=\dfrac{1}{8} \cdot \left(\dfrac{7}{8} \right )^6\sum_{g=7}^{\infty} \cdot \left(\dfrac{7}{8} \right )^{g-7}$

                                                                                                        $\\=\dfrac{1}{8} \cdot \left(\dfrac{7}{8} \right )^6\sum_{g=1}^{\infty} \cdot \left(\dfrac{7}{8} \right )^{g-1} \\ =\dfrac{1}{8}\cdot \left(\dfrac{7}{8} \right )^6 \cdot \dfrac{1}{1-\dfrac{7}{8}}=\left(\dfrac{7}{8} \right )^6$

$P(G \geq 10)=\sum_{g=10}^{\infty}P(G=g)=\sum_{g=10}^{\infty}\dfrac{1}{8} \cdot \left(\dfrac{7}{8} \right )^{g-1}=\dfrac{1}{8} \cdot \left(\dfrac{7}{8} \right )^9\sum_{g=10}^{\infty} \cdot \left(\dfrac{7}{8} \right )^{g-10}$

                                                                                                        $\\=\dfrac{1}{8} \cdot \left(\dfrac{7}{8} \right )^9\sum_{g=1}^{\infty} \cdot \left(\dfrac{7}{8} \right )^{g-1} \\ =\dfrac{1}{8}\cdot \left(\dfrac{7}{8} \right )^9 \cdot \dfrac{1}{1-\dfrac{7}{8}}=\left(\dfrac{7}{8} \right )^9$

$P(G \geq 10|G >6)\dfrac{\left(\dfrac{7}{8} \right )^9}{\left(\dfrac{7}{8} \right )^6}=\left(\dfrac{7}{8} \right )^3=P(G \geq 4)$. Hence Proved.