Question

A baker blends 600 raisins and 400 chocolate chips into a dough mix and, from this, makes 500 cookies. (a) Find the probability that a randomly picked cookie will have no raisins. (b) Find the probability that a randomly picked cookie will have exactly two chocolate chips. (c) Find the probability that a randomly chosen cookie will have at least two bits (raisins or chips) in it.

Answer 1

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