Given that the number of hurricanes follows a Poisson distribution with a mean ($\lambda$) of 5.6 hurricanes per year, we are asked to find the probability that there will be 3 hurricanes in a given year.

The Poisson probability mass function (PMF) is given by:

$P(X = k) = \frac{\lambda^k e^{-\lambda}}{k!}$

where:

• $\lambda = 5.6$ (the mean number of hurricanes per year),
• $k = 3$ (the desired number of hurricanes),
• $e$ is Euler's number (approximately 2.71828).

Step-by-step Solution:

1. Substitute the values $\lambda = 5.6$ and $k = 3$ into the Poisson formula:

$P(X = 3) = \frac{5.6^3 e^{-5.6}}{3!}$

2. Calculate $5.6^3$:

$5.6^3 = 175.616$

3. Compute $e^{-5.6}$:

$e^{-5.6} \approx 0.003698$

4. Find $3!$ (factorial of 3):

$3! = 3 \times 2 \times 1 = 6$

5. Now, plug everything into the formula:

$P(X = 3) = \frac{175.616 \times 0.003698}{6} \approx \frac{0.6491}{6} \approx 0.1082$

Final Answer:

The probability that there will be exactly 3 hurricanes in a year is approximately 0.1082.

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Would you like further details on this, or have any additional questions?

Here are 5 related questions:

1. How would the probability change if we wanted to know the chance of 4 hurricanes?
2. What is the cumulative probability of having fewer than 3 hurricanes in a year?
3. How can we use the Poisson distribution to model events other than hurricanes?
4. If the mean number of hurricanes changed to 6 per year, how would that affect the probability of having exactly 3 hurricanes?
5. How would you use the Poisson distribution to calculate the probability of no hurricanes in a year?

Tip: When calculating probabilities with Poisson distribution, the mean $\lambda$ plays a key role in determining how spread out the possible values are around this mean.