Answer:
a) P(X = 0) for raisins = 0.33287
b) P(X = 2) for chocolate chips = 0.14379
c) P(X ≥ 2) for both bits = 0.59399
Step-by-step explanation:
The average amount of raisin per cookie is 600/500 = 1.2
The average amount of chocolate chips per cookie = 400/500 = 0.8
a) Using Poisson's distribution function
P(X = x) = (e^-λ)(λˣ)/x!
For raisin, Mean = λ = 1.1
x = 0
P(X = 0) = (e⁻¹•¹)(1.1⁰)/(0!) = 0.33287
b) Using Poisson's distribution function
P(X = x) = (e^-λ)(λˣ)/x!
For chocolate chips, Mean = λ = 0.8
x = 2
P(X = 2) = (e⁻⁰•⁸)(0.8²)/(2!) = 0.14379
c) the probability that a randomly chosen cookie will have at least two bits (raisins or chips) in it.
For this, the average number of bit in a raisin = (600+400)/500 = 2 bits per raisin.
P(X ≥ 2) = 1 - [P(X = 0) + P(X = 1)]
For P(X = 0)
P(X = x) = (e^-λ)(λˣ)/x!
Mean = λ = 2
x = 0
P(X = 0) = (e⁻²)(2⁰)/(0!) = 0.13534
For P(X = 1)
P(X = x) = (e^-λ)(λˣ)/x!
Mean = λ = 2
x = 1
P(X = 1) = (e⁻²)(2¹)/(1!) = 0.27067
P(X ≥ 2) = 1 - [P(X = 0) + P(X = 1)]
P(X ≥ 2) = 1 - (0.13534 + 0.27067) = 1 - 0.40601 = 0.59399
