Question 1194140
**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.**