The waiting time T between successive occurrences of an even

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