Help with mgf and probability.?

Note that e^(-3 + 3e^t) is the mgf for the Poisson Distribution with λ = 3,

and (2/3 + (1/3)e^t)^4 is the mgf for the Binomial Distribution with p = 1/3 and n = 4.

Since M_w(t) is the product of these distributions, we must have by uniqueness of mfgs that

W = X + Y, where X ~ Poisson(3) and Y ~ Bin(4, 1/3) are independent RVs.

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X has distribution function P(X = k) = 3^k * e^(-3)/k! for k = 0, 1, 2, ...

Y has distribution function P(Y = j) = C(4, j) (1/3)^j (3/4)^(4-j) for j = 0, 1, 2, 3, 4.

Hence, W = X + Y has distribution (since X and Y are independent)

P(W = k+j with X = k and Y = j) = [3^k * e^(-3)/k!] * [C(4, j) (1/3)^j (3/4)^(4-j)].

Finally, P(W ≤ 1)

= P(W = 0) + P(W = 1)

= P(X = 0, Y = 0) + [P(X = 0, Y = 1) + P(X = 1, Y = 0)]

= (3^0 * e^(-3)/0!) * (C(4, 0) (1/3)^0 (3/4)^4) + (3^0 * e^(-3)/0!) * (C(4, 1) (1/3)^1 (3/4)^3)

+ (3^1 * e^(-3)/1!) * (C(4, 0) (1/3)^0 (3/4)^4)

= e^(-3) * (3/4)^4 + e^(-3) * 4(1/3)(3/4)^3 + 3e^(-3) * (3/4)^4

≈ 0.091.

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I hope this helps!