Question

Ellen catches fish at times of a poisson process with rate 2 per hour. forty percent of the fish are salmon, while 60% of the fish are trout. what is the probability she will catch exactly one salmon and two trout if she fishes for 2.5 h?

Answer 1

To calculate the probability of catching exactly one salmon and two trout in 2.5 hours, we can use the Poisson distribution. The average rate of fish caught per hour is 2, so the average number of fish caught in 2.5 hours is 5. The probability of catching exactly one salmon and two trout can be calculated using the Poisson distribution formula.

Step-by-step explanation:

To calculate the probability of catching exactly one salmon and two trout in 2.5 hours, we need to use the Poisson distribution. The Poisson distribution is used to model the number of events that occur within a fixed interval of time or space, given a known average rate.

The average rate of fish caught per hour is 2, which means that on average, Ellen catches 2 fish per hour. Therefore, the average number of fish caught in 2.5 hours is 2 * 2.5 = 5 fish.

The probability of catching exactly one salmon and two trout can be calculated using the Poisson distribution formula:

P(x) = (e^-λ * λ^x) / x!

where λ is the average rate and x is the number of events.

In this case, λ = 5 and x = 1 (for the salmon) and x = 2 (for the trout).

So, the probability of catching exactly one salmon and two trout in 2.5 hours is:

P(1 salmon and 2 trout) = (e^-5 * 5^1) / 1! * (e^-5 * 5^2) / 2!

Simplifying the equation gives:

P(1 salmon and 2 trout) = (e^-5 * 5 * 10) / 1 * (e^-5 * 5^2 * 10) / 2

P(1 salmon and 2 trout) = (0.00674 * 50) / 1 * (0.00674 * 25) / 2

P(1 salmon and 2 trout) = 0.337 / 2.684

P(1 salmon and 2 trout) ≈ 0.1256 or 12.56%

Answer 2

The chance of Ellen catching a Salmon in an hour is 2 X .40 which is .8 per hour. The chance of Ellen catching a Trout is 2 X .60 which is 1.2 per hour. In 2 and a half hours, Ellen catches 1 Salmon and 2 Trouts is 12.67.