Final answer:
The winning number representing a pizza box with a coupon could arbitrarily be '1' and the non-winning numbers would be 2, 3, and 4. To find the experimental probability of getting the first coupon on the fourth pizza, the calculation is (3/4)^3 * (1/4), which equals 0.4219 after rounding to four decimal places.
Step-by-step explanation:
To determine the experimental probability that a customer would need to buy exactly 4 pizzas before finding a coupon, given that a pizza parlor places coupons in 25 percent of its pizza boxes, we can set up the scenario using a basic probability model. Since 25 percent of the pizza boxes contain a coupon, this means that each box has a 1 in 4 chance of containing a coupon, which can be represented by the numbers 1 through 4, where one of these numbers indicates a winning box with a coupon.
a. If we represent the scenario with the whole numbers 1-4, the winning number could be any one of these numbers. We can arbitrarily decide that the number '1' represents a winning box (since each number has an equal chance of being the winning one).
b. Consequently, the non-winning numbers would be 2, 3, and 4, representing the boxes without coupons.
To find the probability that the first occurrence of finding a coupon is on the fourth pizza box, we have to consider that the first three boxes do not contain a coupon and the fourth one does. This is a sequence of three failures followed by one success.
The probability of not getting a coupon in a single box (q) is 3/4 (or 75 percent), and the probability of getting a coupon (p) is 1/4 (or 25 percent). Using the geometric distribution, the probability of finding the first coupon on the fourth box is:
P(X=4) = q3 * p
P(X=4) = (3/4)3 * (1/4)
P(X=4) = 0.4219 (rounded to four decimal places)