Question

# In a 24 hours period, the average number of boats arriving at a port is 10. Assuming that boats arrive at a random rate that is the same for all subintervals of equal length $i.e. the probability of a boat arriving during a 1 hour period the same for every 1 hour period no matter what$. Calculate the probability that more than 1 boat will arrive during a 1 hour period. $P(X>1$ ) Give your answers to 4 decimal places and in a range between 0 and 1

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## Answer to a math question In a 24 hours period, the average number of boats arriving at a port is 10. Assuming that boats arrive at a random rate that is the same for all subintervals of equal length $i.e. the probability of a boat arriving during a 1 hour period the same for every 1 hour period no matter what$. Calculate the probability that more than 1 boat will arrive during a 1 hour period. $P(X>1$ ) Give your answers to 4 decimal places and in a range between 0 and 1

Fred
4.4
To solve this problem, we will use the Poisson distribution since we are dealing with the arrival of boats over a given time period.

The average number of boats arriving at the port in a 1-hour period is given as 10. Therefore, the average rate parameter, λ, is also 10.

The probability of more than 1 boat arriving during a 1-hour period can be calculated as:

P$X > 1$ = 1 - P$X = 0$ - P$X = 1$

where P$X = k$ represents the probability of k boats arriving during a 1-hour period.

Using the formula for the Poisson distribution:

P$X = k$ = $e^(-λ$ * λ^k) / k!

we can calculate each term.

P$X = 0$ = $e^(-10$ * 10^0) / 0! = e^$-10$

P$X = 1$ = $e^(-10$ * 10^1) / 1! = 10 * e^$-10$

Now we can substitute these values into the formula for P$X > 1$:

P$X > 1$ = 1 - e^$-10$ - 10 * e^$-10$

Calculating this expression, we find:

P$X > 1$ ≈ 1 - e^$-10$ - 10 * e^$-10$ ≈ 1 - 0.00004540 - 0.00045399 ≈ 0.9995

Therefore, the probability that more than 1 boat will arrive during a 1-hour period is approximately 0.9995.

\textbf{Answer: } P$X > 1$ \approx 0.9995

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