[math: P(k events in interval t)=(rt)ke−rtk!]P(k ext{ events in interval }t) = \frac{(rt)^{k}e^{-rt}}{k!}Suppose that earthquakes occurs in the western of the United States are approximately Poisson at a rate of [math: r=2] per week. Find the probability that at least 3 earthquakes occur during the next 2 weeks.

Question

[math: P(k events in interval t)=(rt)ke−rtk!]P(k	ext{ events in interval }t) = \frac{(rt)^{k}e^{-rt}}{k!}Suppose that earthquakes occurs in the western of the United States are approximately Poisson at a rate of [math: r=2] per week. Find the probability that at least 3 earthquakes occur during the next 2 weeks.

BlackTom AI · Accepted Answer

We start by restating the core problem in our own terms: earthquakes in the western U.S. follow a Poisson process with rate r = 2 per week, and we want the probability that at least 3 earthquakes occur during the next 2 weeks.

Step 1: determine the Poisson parameter for the interval of interest. For a Poisson process, the expected number of events in a time interval of length t is λ = r t. Here t = 2 weeks and r = 2 per week, so λ = 2 × 2 = 4.

Step 2: express the probability of at least 3 events. The Poisson probability mass function is P(K = k) = (λ^k e^{-λ}) / k!. The probability of at least 3 events is P(K ≥ 3) = 1 − [P(0) + P(1) + P(2)].

Step 3: compute P(0), P(1), P(2) with λ = 4.
- P(0) = (4^0 e^{-4}) / 0! = e^{-4} ≈ 0.018315.
- P(1) = (4^1 e^{-4}) / 1! = 4 e^{-4} ≈ 0.073260.
- P(2) = (4^2 e^{-4}) / 2! = 16 e^{-4} / 2 ≈ 8 e^{-4} ≈ 0.146520.

Step 4: sum the probability for 0, 1, or 2 events and subtract from 1.
- P(0) + P(1) + P(2) ≈ 0.018315 + 0.073260 + 0.146520 ≈ 0.238095.
- Therefore P(K ≥ 3) = 1 − 0.238095 ≈ 0.761905.

Step 5: compare with the provided options. The computed probability rounds to 0.7619, which corresponds to the option labeled with that value in the list.

Notes on other options:
- If you look at P(K ≥ 3) as 1 minus the mass up to k = 2, any error in calculating e^{-4} or in computing the powers 4^k can lead to markedly different results (for example, miscomputing P(2) as 0.07326 instead of 0.14652 would yield a much smaller final probability).
- Option a (None of These) would be incorrect if the calculation yields a finite, matching value among the listed choices.
- Option c (0.1465) would correspond to a single term like P(2) alone, not the cumulative probability for at least 3 events.
- Option d (0.6767) and option e (0.3233) would require different cumulative contributions from P(0), P(1), and P(2), which do not align with the correct Poisson tail for λ = 4.

In summary, the method uses the Poisson distribution with λ = rt = 4, computes the probability of 0, 1, or 2 events, subtracts from 1, and obtains approximately 0.7619, matching the given option value.

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