Question

a class orders a variety of pizzas for their end-of-year party. they order 2 cheese, 3 pepperoni, 1 supreme, and 1 veggie. the teacher invites small group representatives, one at a time, to select a box of pizza to share with their small group. what is the probability that the first representative selects a cheese pizza and the second representative selects a pepperoni pizza?

a) $\frac{6}{49}$

b) $\frac{1}{7}$

c) $\frac{2}{7}$

d) $\frac{5}{7}$

Explanation:

Step1: Calculate total number of pizzas

First, find the total number of pizzas by adding the number of each type: \(2 + 3 + 1 + 1 = 7\) pizzas.

Step2: Probability of first selecting cheese

The number of cheese pizzas is 2, so the probability that the first representative selects a cheese pizza is \(\frac{2}{7}\) (since there are 2 cheese out of 7 total).

Step3: Probability of second selecting pepperoni (after first selection)

After one cheese pizza is selected, there are now \(7 - 1 = 6\) pizzas left. The number of pepperoni pizzas is 3, so the probability that the second representative selects a pepperoni pizza is \(\frac{3}{6}=\frac{1}{2}\).

Step4: Multiply the two probabilities

To find the probability of both events happening (first cheese, then pepperoni), multiply the two probabilities: \(\frac{2}{7} \times \frac{3}{6}=\frac{2}{7} \times \frac{1}{2}=\frac{1}{7}\).

Answer:

b) \(\frac{1}{7}\)