Probability of P(X = Y) with Poisson and Geometric distributions

Question

Random independent variables X and Y have distributions of X ~ P(λ) (Poisson distribution) and Y ~ G(p) (geometric distribution), λ = 7.3, p = 0.6. Find the probability P(X = Y).

Answer 1

$$q = 1 - 0.6 = 0.4$$

$$P(X = k) = \frac{λ^k}{k!} ⋅ e^{-λ}, k = 0, 1, 2...$$

$$P(Y = k) = q^{k - 1}⋅p, k = 1, 2, 3...$$

$$P(X = Y) = P(X = Y = 1) + P(X = Y = 2) + {...}$$

$$P(X = Y) = P(X = 1)⋅P(Y=1) + P(X = 2)⋅P(Y=2) + {...}$$

$$P(X = Y) = \sum_{k = 1}^∞\frac{λ^k}{k!} ⋅ e^{-λ}⋅q^{k - 1}⋅p = e^{-λ}⋅p ⋅\sum_{k = 1}^∞\frac{λ^k}{k!} ⋅q^{k - 1}$$

$$P(X = Y) = e^{-λ}⋅p ⋅\sum_{k = 1}^∞\frac{λ^k}{k!} ⋅q^{k - 1} = e^{-λ}⋅p ⋅\frac{1}{q}⋅\sum_{k = 1}^∞\frac{(λq)^k}{k!}$$

$$P(X = Y) = e^{-λ}⋅p ⋅\frac{1}{q} ⋅ (\frac{λq}{1!} + \frac{(λq)^2}{2!} + \frac{(λq)^3}{3!} + {...})$$

$$P(X = Y) = e^{-λ}⋅p ⋅\frac{1}{q} ⋅ (e^{λq} - 1)$$

$$P(X = Y) = e^{-6.3}⋅0.6 ⋅\frac{1}{0.4} ⋅ (e^{6.3{⋅}0.4} - 1) ≈ 0.03148 $$

Thanks to @lulu for the help with this problem!