The waiting time T between consecutive occurrences of an eve

The waiting time T between consecutive occurrences of an event E in a discrete-time renewal process has the probability P(T= 1)=0.5, P(T= 2) = 0.5 Denoting by W5 the waiting time to the fifth event, find the probability P(W5 = 7). (Correct to four decimal places).

Solution

Let the waiting time to 1st occurence be T₁ , between occurence 1 and 2 be T₂ , and.... between occurence 4 and 5 be T₅.

P(T₁ + T₂ + T₃ + T₄ + T₅ = 7) :

For the 5 time intervals to add up to 7, as the only possible interval values are 1 and 2, 2 of the intervals have to have the value 2 and remaining 3 have to have the value 1

2+2+1+1+1 = 7

These values can be assigned to 5 intervals in : 5C2 * 3C3 ways = 10 ways

P(T₁ + T₂ + T₃ + T₄ + T₅ = 7) = 10 * [P(T=1)]³ * [P(T=2)]² = 10*0.5³ *0.5² = 0.3125