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 0.152 . Consider your next 27 purchases that produce a game piece. Calculate the following: Round your answers to at least three decimal places. a) What is the probability that you win 4 prizes? 0.2112 $\sigma^{6}$ b) What is the probability that you win more than 6 prizes? c) What is the probability that you win between 3 and 5 (inclusive) prizes? d) What is the probability that you win 3 prizes or fewer? Question Help: $D$ Post to forum

Step 1 ：This problem is about the binomial distribution. The binomial distribution model deals with finding the probability of success of an event which has only two possible outcomes in a series of experiments. For a random variable X if n is the number of experiments, p is the probability of success in a single experiment, then the probability mass function of X is given by: \(P(X=k) = C(n, k) * (p^k) * ((1-p)^(n-k))\), where \(C(n, k)\) is the combination of n items taken k at a time.

Step 2 ：In this case, \(n=27\) (the number of purchases), \(p=0.152\) (the probability of winning a prize), and we want to find \(P(X=4)\), the probability of winning 4 prizes.

Step 3 ：Substituting the given values into the formula, we get: \(P(X=4) = C(27, 4) * (0.152^4) * ((1-0.152)^(27-4))\).

Step 4 ：Calculating the above expression, we get a probability of approximately 0.211.

Step 5 ：Final Answer: The probability that you win 4 prizes is approximately \(\boxed{0.211}\).