Question 1194140: X and Y are independent, X ~ P(λ1), Y ~ P(λ2). Given probabilities P(X1 > 0) = 0.58, P(X2 > 0) = 0.43. Find the probability of P(X+Y=1)
Answer by parmen(42)   (Show Source):
You can put this solution on YOUR website! **1. Determine λ1 and λ2:**
* We know that for a Poisson distribution, P(X > 0) = 1 - P(X = 0)
* Since X ~ P(λ1), P(X = 0) = e^(-λ1)
* Therefore, P(X > 0) = 1 - e^(-λ1)
* Given P(X > 0) = 0.58, we can solve for λ1:
0.58 = 1 - e^(-λ1)
e^(-λ1) = 0.42
λ1 = -ln(0.42) ≈ 0.8675
* Similarly, for Y ~ P(λ2):
P(Y > 0) = 1 - e^(-λ2)
Given P(Y > 0) = 0.43:
0.43 = 1 - e^(-λ2)
e^(-λ2) = 0.57
λ2 = -ln(0.57) ≈ 0.5621
**2. Calculate P(X = 0) and P(Y = 0):**
* P(X = 0) = e^(-λ1) = e^(-0.8675) ≈ 0.42
* P(Y = 0) = e^(-λ2) = e^(-0.5621) ≈ 0.57
**3. Calculate P(X = 1) and P(Y = 1):**
* For a Poisson distribution, P(X = k) = (e^(-λ) * λ^k) / k!
* P(X = 1) = (e^(-λ1) * λ1^1) / 1! = e^(-0.8675) * 0.8675 ≈ 0.3698
* P(Y = 1) = (e^(-λ2) * λ2^1) / 1! = e^(-0.5621) * 0.5621 ≈ 0.3155
**4. Calculate P(X + Y = 1):**
* P(X + Y = 1) can occur in two ways:
* X = 1 and Y = 0
* X = 0 and Y = 1
* Since X and Y are independent:
P(X + Y = 1) = P(X = 1) * P(Y = 0) + P(X = 0) * P(Y = 1)
P(X + Y = 1) = 0.3698 * 0.57 + 0.42 * 0.3155 ≈ 0.3423
**Therefore, the probability of P(X + Y = 1) is approximately 0.3423.**
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