The waiting time T between successive occurrences of an event E in a discrete-time renewal process has the probability distribution P(T = 1) = 0.7, P(T = 2)- 0.3. The waiting time to the sixth occurrence of E is denoted by W6. Find the probability P(W6 = 10). 0.0595 0.0494 0.0393 Solution Let the waiting time to 1st occurence be T1 , between occurence 1 and 2 be T2 , and.... between occurence 5 and 6 be T6. P(T1 + T2 + T3 + T4 + T5 + T6 = 10) : For the 6 time intervals to add up to 10, as the only possible interval values are 1 and 2, 4 of the intervals have to have the value 2 and remaining 2 have to have the value 1 2+2+2+2+1+1 = 10 These values can be assigned to 6 intervals in : 6C4 * 2C2 ways = 15 ways P(T1 + T2 + T3 + T4 + T5 + T6 = 10) = 15 * [P(T=1)]2 * [P(T=2)]4 = 15*0.72 *0.34 = 0.0595 Answer: 0.0595.

1. The waiting time T between successive occurrences of an event E in a discrete-time renewal
process has the probability distribution P(T = 1) = 0.7, P(T = 2)- 0.3. The waiting time to the
sixth occurrence of E is denoted by W6. Find the probability P(W6 = 10). 0.0595 0.0494
0.0393
Solution
Let the waiting time to 1st occurence be T1 , between occurence 1 and 2 be T2 , and.... between
occurence 5 and 6 be T6.
P(T1 + T2 + T3 + T4 + T5 + T6 = 10) :
For the 6 time intervals to add up to 10, as the only possible interval values are 1 and 2, 4 of the
intervals have to have the value 2 and remaining 2 have to have the value 1
2+2+2+2+1+1 = 10
These values can be assigned to 6 intervals in : 6C4 * 2C2 ways = 15 ways
P(T1 + T2 + T3 + T4 + T5 + T6 = 10) = 15 * [P(T=1)]2 * [P(T=2)]4 = 15*0.72 *0.34 = 0.0595
Answer: 0.0595