Question

The chance of having an extra fortune in a fortune cookie is about 3%. Given a bag of 144 fortune cookies, we are interested in the number of cookies with an extra fortune. Two distributions may be used to solve this problem, but only use one distribution to solve the problem.

a. In words, define the random variable X.

b. List the values that X may take on.

c. Give the distribution of X. X ~ _____(_____,_____)

d. How many cookies do we expect to have an extra fortune?

e. Find the probability that none of the cookies have an extra fortune.

f. Find the probability that more than three have an extra fortune.

g. As n increases, what happens involving the probabilities using the two distributions? Explain in complete sentences.

Short answer

a. The random variable X is the number of fortune cookies that have an extra fortune.

b. The values of X are $0,1,2,3,...........,144$

c. The distribution of X are

d. The expected number of cookies that have an extra fortune is $4.32$

e. The probability that none of the cookies have an extra fortune is$0.0124or0.0133$

f. The probability that more than three have an extra fortune is localid="1649245875557" $0.6300or0.6264$

g. As n increases the probabilities come close together.

Step-by-step solution

Content Introduction

In a large population, the Poisson distribution is used to characterize the distribution of unusual events.

Explanation (part a)

We are given,

A bag of $144$fortune cookies and the chance of having an extra fortune in a fortune cookie is $3\%$. Random variable in simple terms generally refers to variables whose values are unknown, therefore, in this case X is the number of fortune cookies that have an extra fortune.

Explanation (part b)

The list of values that we use for values of X are$0,1,2,3,.....,144.$

Explanation (part c)

Here, the random variable X refers to the number of trials. Each trial is independent of others and has similar probability of success. This implies that random variable X follows Poisson Distribution.

The Poisson distribution is shown as$X~P(x,\mu )$where x is the successes occurs in poisson distribution and$\mu$is the mean successes.

Thus, the distribution of X is role="math" localid="1649244224297" $X~P(4.32)$

Explanation (part d)

We are given, that in a bag of $144$fortune cookies there is a $0.03$chance of having an extra fortune cookie.

The expected number of cookies that have an extra fortune is calculated as follow:

$$\begin{gathered} X=144\times 0.03 \\ X=4.32 \end{gathered}$$

Explanation (part e)

The probability in poisson distribution is as follow:

$f\left(x\right)=\frac{{\lambda }_x}{x!}{e}^{-\lambda }$

where, $\lambda =4.32,x=0$

localid="1649246766111" $$\begin{gathered} P_n(X=0)=\frac{4.{32}^0}{0!}{e}^{-4.32} \\ {P}_n(X=0)=0.0124or0.0133 \end{gathered}$$

Explanation (part f)

The probability that more than three have an extra fortune is calculated as follow:

$$\begin{gathered} P_n(\xi \le 3)=P_n\left(0\right)+P_n\left(1\right)+P_n\left(2\right)+P_n\left(3\right) \\ {P}_n(\xi \le 3)=0.00335+0.01907+0.05436+0.01327 \\ {P}_n(\xi \le 3)=0.6300or0.6264 \end{gathered}$$

Explanation (part g)

As n increases the probabilities comes closer together.