Question

Aja's favorite cereal is running a promotion that says 111-in-444 boxes of the cereal contain a prize. Suppose that Aja is going to buy 555 boxes of this cereal, and let XXX represent the number of prizes she wins in these boxes. Assume that these boxes represent a random sample, and assume that prizes are independent between boxes. What is the probability that she wins at most 111 prize in the 555 boxes

Answer 1

Using the binomial distribution, there is a 0.6328 = 63.28% probability that she wins at most 1 prize.

For each box, there are only two possible outcomes, either it has a prize, or it does not. The probability of a box having a prize is independent of any other box, hence, the binomial distribution is used to solve this question.

Binomial probability distribution

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$C_{n,x} = \frac{n!}{x!(n-x)!}$

The parameters are:

• x is the number of successes.

• n is the number of trials.

• p is the probability of a success on a single trial.

In this problem:

• She buys 5 boxes, hence $n = 5$

• 1 in 4 boxes has a prize, hence $p = \frac{1}{4} = 0.25$

The probability is:

$P(X \leq 1) = P(X = 0) + P(X = 1)$

Hence:

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{5,0}.(0.25)^{0}.(0.75)^{5} = 0.2373$

$P(X = 1) = C_{5,1}.(0.25)^{1}.(0.75)^{4} = 0.3955$

Then

$P(X \leq 1) = P(X = 0) + P(X = 1) = 0.2373 + 0.3955 = 0.6328$

0.6328 = 63.28% probability that she wins at most 1 prize.

A similar problem is given at brainly.com/question/24863377