A die is rolled until the first time T that a six turns up. (a) What is the probability distribution for T?
Find P(T > 3).
Find P(T > 6|T > 3).
First time T with six turn up: \[ T = \tfrac{1}{6} \]
Distribution Formula: \[ P(T = n) = q^{n-1}p \\ where \ q = 1 - p \]
Ans. # A) What is the probability distribution for T?
\[ P(n) = q^{n-1}p \\ P(n) = \big(1 - \tfrac{1}{6 } \big)^{n-1} *\ \tfrac{1}{6} \\ P(n) = \big( \tfrac{6}{6} - \tfrac{1}{6} \big)^{n-1} \ * \tfrac{1}{6} \\ P(n) = \big( \tfrac{5}{6} \big)^{n-1} *\ \tfrac{1}{6} \]
Ans # b) Find P(T > 3)
\[ P \big( T > K \big) = q^{k} => \big(1 - p\big)^k \\ P \big(T > 3 \big) = \big(1 - \tfrac{1}{6} \big) => \big( \tfrac{6}{6} - \tfrac{1}{6}\big)^3 \\ P \big( T > 3 \big) = \big(\tfrac{5}{6} \big)^3 = \tfrac{125}{216} \]
Ans # c) Find P(T > 6|T > 3)
\[ P \big(T >6 | T > 3 \big) = \big(1 - p\big)^3 \\ P \big(T > 6 | T > 3 \big) = \big(1 - \tfrac{1}{6}\big)^3 \\ P \big(T > 6 | T > 3 \big) = \big(\tfrac{6}{6} - \tfrac{1}{6}\big)^3 \\ P \big(T > 6 | T > 3 \big) = \big(\tfrac{5}{6}\big)^3 \\ P \big(T > 6 | T > 3 \big) = \tfrac{125}{216} \]