The probability that an event A occurs during an interval of…
The probability that an event $A$ occurs during an interval of length $T$ is $p$, and the event is equally likely to occur at any moment of this interval. It is known that during the time $t\lt T$ the event has not occurred. Determine the probability that the event $A$ will occur in the remaining part of the interval.
Explanation of Method
"Equally likely at any moment" means: the probability that $A$ occurs in a subinterval of length $s$ is $p\frac sT$. Let $B$ = "$A$ does not occur in $[0,t]$" and $C$ = "$A$ occurs in $(t,T]$". We need $P(C\mid B)=\frac{P(BC)}{P(B)}$. Since $C\subseteq B$ ($A$ occurs at most once), $P(BC)=P(C)$.
Step 1: The probabilities of the pieces
$A$ occurs somewhere in $[0,T]$ with probability $p$, uniformly in time, so
\[ P(A \text{ in } [0,t])=p\frac tT, \qquad P(A\text{ in }(t,T])=p\frac{T-t}{T}. \]
Step 2: The condition
\[ P(B)=P(A\text{ not in }[0,t])=1-p\frac tT. \]
Step 3: Conditional probability
If $A$ occurs in $(t,T]$, it has certainly not occurred in $[0,t]$, so $BC=C$:
\[ \boxed{P(C\mid B)=\frac{p\,\frac{T-t}{T}}{1-p\,\frac tT}=\frac{p\,(T-t)}{T-pt}.} \]
Step 4: Checks
- $t=0$: $P=p$ (no information yet) ✓.
- $t\to T$: $P\to0$ (no time left) ✓.
- $p=1$: $P=1$ (the event is certain to occur, and it has not occurred yet, so it must occur later) ✓.
Step 5: Example
A parcel arrives during the day ($T=10$ hours) with probability $p=0.8$, at a uniformly random time. After $t=5$ hours it has not arrived:
\[ P=\frac{0.8\cdot5}{10-4}=\frac{4}{6}\approx0.667. \]
The unconditional probability for the second half of the day was only $0.4$; not having received it yet makes a later arrival more likely.
Result
| Answer | |
|---|---|
| Probability | $\dfrac{p(T-t)}{T-pt}=\dfrac{p(1-t/T)}{1-pt/T}$ |