Problem setup (statistics context)
Let x represent the number of minutes passed. Consider a counting variable \(N(x)\) that records how many events occur during the time interval of length \(x\) minutes. A standard statistical model for counts over time is a Poisson process, characterized by:
- a constant average rate \(\lambda\) events per minute,
- independent counts in disjoint time intervals (independent increments),
- the expected count proportional to time.
Assumed rate: \(\lambda = 0.5\) events per minute.
Over \(x\) minutes, the expected number of events is \[ \mu(x) = \lambda x. \]
Step 1: Identify the distribution of \(N(x)\)
Under a Poisson counting model, the number of events in \(x\) minutes follows: \[ N(x) \sim \text{Poisson}(\lambda x). \]
Step 2: Write the probability of exactly k events in x minutes
The probability mass function for \(k=0,1,2,\dots\) is: \[ P\!\left(N(x)=k\right) = e^{-\lambda x}\cdot\frac{(\lambda x)^k}{k!}. \]
Step 3: Compute example probabilities for a specific time (x = 10 minutes)
Take \(x=10\) minutes. First compute the mean count: \[ \mu = \lambda x = 0.5 \times 10 = 5. \]
3A. Probability of exactly 3 events in 10 minutes
Use \(k=3\) and \(\mu=5\): \[ P(N(10)=3) = e^{-5}\cdot\frac{5^3}{3!}. \]
Compute the factorial and power: \[ 3! = 6,\qquad 5^3 = 125. \]
Substitute: \[ P(N(10)=3) = e^{-5}\cdot\frac{125}{6}. \]
Numerically, \[ e^{-5}\approx 0.00673795,\qquad \frac{125}{6}\approx 20.83333, \] so \[ P(N(10)=3)\approx 0.00673795 \times 20.83333 \approx 0.14037. \]
3B. Probability of at least 1 event in 10 minutes
Use the complement rule: \[ P(N(10)\ge 1)=1-P(N(10)=0). \]
For \(k=0\), \[ P(N(10)=0)=e^{-5}\cdot\frac{5^0}{0!}=e^{-5}\cdot\frac{1}{1}=e^{-5}\approx 0.00673795. \]
Therefore, \[ P(N(10)\ge 1)\approx 1-0.00673795=0.99326. \]
| Quantity | Exact expression | Approx. value |
|---|---|---|
| \(\mu\) for \(x=10\) | \(\mu=\lambda x=0.5\times 10\) | \(5\) |
| \(P(N(10)=3)\) | \(e^{-5}\cdot\frac{5^3}{3!}\) | \(\approx 0.14037\) |
| \(P(N(10)\ge 1)\) | \(1-e^{-5}\) | \(\approx 0.99326\) |
Step 4: Solve for x when the probability of zero events is 0.05
The probability of zero events in \(x\) minutes is: \[ P(N(x)=0)=e^{-\lambda x}. \]
Set this equal to \(0.05\) and solve: \[ e^{-0.5x}=0.05. \]
Take natural logarithms: \[ -0.5x=\ln(0.05). \]
Divide by \(-0.5\): \[ x=\frac{-\ln(0.05)}{0.5}. \]
Using \(\ln(0.05)\approx -2.99573\), \[ x\approx \frac{2.99573}{0.5}=5.99146. \]
Result: \(x \approx 5.99\) minutes makes \(P(N(x)=0)=0.05\) when \(\lambda=0.5\) events per minute.
Visualization: Poisson probabilities when x = 10 minutes (\(\mu=5\))
Conclusion
Letting x represent the number of minutes passed leads to a time-scaled Poisson model \(N(x)\sim\text{Poisson}(\lambda x)\), enabling exact probability calculations for counts over time and direct solving for \(x\) from target probabilities such as \(P(N(x)=0)=0.05\).