The problem is a classic example of a Poisson distribution problem. The Poisson distribution gives the probability of a given number of events (in this case, road accidents) happening in a fixed interval of time or space (in this case, a month) if these events occur with a known average rate (in this case, 7 accidents per month) and independently of the time since the last event.

The formula for the Poisson probability is:

P(k; λ) = λ^k * e^-λ / k!

where:
- P(k; λ) is the Poisson probability,
- k is the actual number of successes that result from the experiment,
- λ is the mean number of successes that occur in a specified region,
- e is the number approximately equal to 2.71828 (Euler's number),
- k! is the factorial of k.

In this case, we want to find the probability of observing exactly three accidents next month, so k = 3. The average number of accidents per month is 7, so λ = 7.

Substituting these values into the formula, we get:

P(3; 7) = 7^3 * e^-7 / 3!
= 343 * e^-7 / 6
= 343 * 0.000911882 / 6
= 0.312676 / 6
= 0.0521127

So, the probability of observing exactly three accidents on this stretch of road next month is approximately 0.052, or 5.2%. Therefore, the closest answer among the options given is 0.052.

Question

The problem is a classic example of a Poisson distribution problem. The Poisson distribution gives the probability of a given number of events (in this case, road accidents) happening in a fixed interval of time or space (in this case, a month) if these events occur with a known average rate (in this case, 7 accidents per month) and independently of the time since the last event.

The formula for the Poisson probability is:

P(k; λ) = λ^k * e^-λ / k!

where:
- P(k; λ) is the Poisson probability,
- k is the actual number of successes that result from the experiment,
- λ is the mean number of successes that occur in a specified region,
- e is the number approximately equal to 2.71828 (Euler's number),
- k! is the factorial of k.

In this case, we want to find the probability of observing exactly three accidents next month, so k = 3. The average number of accidents per month is 7, so λ = 7.

Substituting these values into the formula, we get:

P(3; 7) = 7^3 * e^-7 / 3!
= 343 * e^-7 / 6
= 343 * 0.000911882 / 6
= 0.312676 / 6
= 0.0521127

So, the probability of observing exactly three accidents on this stretch of road next month is approximately 0.052, or 5.2%. Therefore, the closest answer among the options given is 0.052.

Knowee AI · Accepted Answer

The problem is a classic example of a Poisson distribution problem. The Poisson distribution gives the probability of a given number of events (in this case, road accidents) happening in a fixed interval of time or space (in this case, a month) if these events occur with a known average rate (in this case, 7 accidents per month) and independently of the time since the last event.

The formula for the Poisson probability is:

P(k; λ) = λ^k * e^-λ / k!

where:
- P(k; λ) is the Poisson probability,
- k is the actual number of successes that result from the experiment,
- λ is the mean number of successes that occur in a specified region,
- e is the number approximately equal to 2.71828 (Euler's number),
- k! is the factorial of k.

In this case, we want to find the probability of observing exactly three accidents next month, so k = 3. The average number of accidents per month is 7, so λ = 7.

Substituting these values into the formula, we get:

P(3; 7) = 7^3 * e^-7 / 3!
= 343 * e^-7 / 6
= 343 * 0.000911882 / 6
= 0.312676 / 6
= 0.0521127

So, the probability of observing exactly three accidents on this stretch of road next month is approximately 0.052, or 5.2%. Therefore, the closest answer among the options given is 0.052.

The average number of road accidents that occur on a particular stretch of road during a month is 7. What is the probability of observing exactly three accidents on this stretch of road next month?0.0200.0230.0480.052

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