Original practice · fully worked
Original variant: memoryless continuation after three failed attempts
Independent connection attempts succeed with probability 0.35. It is known that the first three attempts failed. Find the conditional probability that the first success occurs after attempt five.
- A
0.1225 - B
0.4225 - C
0.5775 - D
0.6500 - E
0.7250
Answer and calculation
After the first three failures, the geometric process restarts. Passing attempt five requires two additional failures, probability 0.65²=0.4225, choice B.
Setup
Setup
Let q=0.65 be the probability that a single connection attempt fails.
q=1−0.35=0.65 Model
Model
Given the first three failures, the geometric memoryless property reduces the event T greater than 5 to failures on attempts 4 and 5.
Pr(T>5∣T>3)=Pr(attempts 4 and 5 both fail) Compute
Compute
Independence of the attempts gives q²=0.65²=0.4225 for the two additional failures.
Pr(T>5∣T>3)=q2=0.4225 Answer
Answer
The conditional probability that the first success occurs after attempt five is 0.4225, which is choice B.
0.4225(B) 