Question

A large fast-food restaurant is having a promotional game where game pieces can be found on various products. Customers can win food or cash prizes. According to the company, the probability of winning a prize (large or small) with any eligible purchase is 0.117. Consider your next 29 purchases that produce a game piece. Round each answer to at least three decimal places. a) What is the probability that you win at least three prizes? b) What is the probability that you win more than six prizes? c) What is the probability that you win one prizes or fewer?

Answer 1

Let's start by calculating the probability of winning no prize with one eligible purchase, which is:

P(no prize) = 1 - 0.117 = 0.883

a) The probability of winning at least three prizes in 29 purchases can be calculated using the binomial distribution:

P(X ≥ 3) = 1 - P(X < 3) = 1 - [P(X = 0) + P(X = 1) + P(X = 2)]

where X is the number of prizes won in 29 purchases. We can use the binomial probability formula to calculate each of these individual probabilities:

P(X = k) = C(n, k) * p^k * (1 - p)^(n - k)

where n = 29 is the number of purchases, p = 0.117 is the probability of winning a prize, and C(n, k) is the number of ways to choose k prizes out of n purchases. Using a calculator or software, we can find:

P(X = 0) ≈ 0.183

P(X = 1) ≈ 0.315

P(X = 2) ≈ 0.267

Therefore,

P(X ≥ 3) ≈ 1 - (0.183 + 0.315 + 0.267) ≈ 0.234

So the probability of winning at least three prizes in 29 purchases is approximately 0.234.

b) The probability of winning more than six prizes can be calculated using a similar method:

P(X > 6) = 1 - P(X ≤ 6) = 1 - [P(X = 0) + P(X = 1) + ... + P(X = 6)]

Using a calculator or software, we can find:

P(X > 6) ≈ 1 - 0.996 ≈ 0.004

So the probability of winning more than six prizes in 29 purchases is approximately 0.004.

c) The probability of winning one prize or fewer can be calculated as:

P(X ≤ 1) = P(X = 0) + P(X = 1)

Using the binomial probability formula, we can find:

P(X = 0) ≈ 0.183

P(X = 1) ≈ 0.315

Therefore,

P(X ≤ 1) ≈ 0.183 + 0.315 ≈ 0.498

So the probability of winning one prize or fewer in 29 purchases is approximately 0.498.

Answer 2

Answer: We can use the binomial distribution to solve this problem. Let X be the number of prizes won in 29 purchases, and p be the probability of winning a prize with any eligible purchase, which is 0.117.

a) The probability of winning at least three prizes is:

P(X ≥ 3) = 1 - P(X < 3)

To find P(X < 3), we can use the binomial cumulative distribution function with n = 29, p = 0.117, and x = 2:

P(X < 3) = F(2; 29, 0.117) = 0.906

Therefore, P(X ≥ 3) = 1 - 0.906 = 0.094 (rounded to three decimal places).

b) The probability of winning more than six prizes is:

P(X > 6) = 1 - P(X ≤ 6)

To find P(X ≤ 6), we can use the binomial cumulative distribution function with n = 29, p = 0.117, and x = 6:

P(X ≤ 6) = F(6; 29, 0.117) = 0.985

Therefore, P(X > 6) = 1 - 0.985 = 0.015 (rounded to three decimal places).

c) The probability of winning one prize or fewer is:

P(X ≤ 1) = F(1; 29, 0.117) = 0.698 (rounded to three decimal places).

Therefore, the answer is 0.698 (rounded to three decimal places).

Explanation: