The number of visible defects on a product container is thought to be Poisson distributed with a mean equal to 3.5 defects per container. Based on this, the probability that 2 containers will contain a total of less than 2 defects is:

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BlackTom AI · Accepted Answer

To tackle this, I’ll first restate what the problem is asking: we have two containers, each with a Poisson-distributed number of defects having mean 3.5, and we want the probability that the total defects across the two containers is less than 2.

Option 1: 0.1850. This value is clearly too large for a Poisson sum with a high mean. When the combined mean is λ = 3.5 + 3.5 = 7 for two containers, the probability of seeing fewer than 2 events should be quite small, because the Poisson distribution with mean 7 places most mass around 7 defects. Therefore 0.1850 overestimates the tail probability at 0 or 1 defect.

Option 2: 0.1359. This is still far larger than the actual tail probability for λ = 7. It suggests a miscalculation or using a much smaller λ or a different distribution altogether. In a Poisson context, the probability of 0 or 1 defect should be around a percent-perhaps less-than-a-percent range, given λ = 7, not around 0.1–0.14.

Option 3: 0.0223. This is closer to the ballpark for a tail probability, but it’s still higher than the true value for P(total defects < 2) when λ = 7. The correct computation involves P(0) + P(1) with P(k) = e^{-λ} λ^k / k!, so using λ = 7 yields a sum around 0.0073, not 0.0223.

Option 4: 0.0073. This matches the exact Poisson calculation for total defects across two containers. Since the total mean is λ = 3.5 + 3.5 = 7, the probability of observing 0 or 1 defect is P(0) + P(1) = e^{-7} (1 + 7) = e^{-7} × 8 ≈ 0.0009119 × 8 ≈ 0.007295, which rounds to 0.0073.

Option 5: 0.0296. Like option 3, this is a plausible tail-area value for some λ, but it does not correspond to the actual P(total < 2) when λ = 7. The precise sum P(0) + P(1) is substantially smaller than 0.0296, making this option incorrect for the given setup.

In summary, the correct approach is to treat the sum of two independent Poisson variables as Poisson with mean equal to the sum of their means, here λ = 7. Then compute P(T < 2) = P(0) + P(1) = e^{-7} (1 + 7) = 8 e^{-7} ≈ 0.0073. The other options either overstate the tail probability or rely on incorrect λ or distribution assumptions.

OSC 2030-009 Fall 2025 Exam 1

The number of visible defects on a product container is thought to be Poisson distributed with a mean equal to 3.5 defects per container. Based on this, the probability that 2 containers will contain a total of less than 2 defects is:

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