Question

A well-mixed cookie dough will produce cookies with a mean of 7 chocolate chips apiece. What is the probability of getting a cookie with at least 3 chips? Round your answer to four decimal places.

Answer 1

Answer:

0.9704 = 97.04% probability of getting a cookie with at least 3 chips

Step-by-step explanation:

We only have the mean number of chocolate chips. So we use the Poisson distribution to solve this question.

Poisson distribution:

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

In which

x is the number of sucesses

e = 2.71828 is the Euler number

$\mu$ is the mean in the given time interval.

A well-mixed cookie dough will produce cookies with a mean of 7 chocolate chips apiece.

This means that $\mu = 7$

What is the probability of getting a cookie with at least 3 chips?

Either we have less than 3 chips, or we have at least 3 chips. The sum of the probabilities of these events is decimal 1.

$P(X < 3) + P(X \geq 3) = 1$

We want $P(X \geq 3)$. Then

$P(X \geq 3) = 1 - P(X < 3)$

In which

$P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2)$

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

$P(X = 0) = \frac{e^{-7}*7^{0}}{(0)!} = 0.0009$

$P(X = 1) = \frac{e^{-7}*7^{1}}{(1)!} = 0.0064$

$P(X = 2) = \frac{e^{-7}*7^{2}}{(2)!} = 0.0223$

$P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) = 0.0009 + 0.0064 + 0.0223 = 0.0296$

$P(X \geq 3) = 1 - P(X < 3) = 1 - 0.0296 = 0.9704$

0.9704 = 97.04% probability of getting a cookie with at least 3 chips