Question

A pizza restaurant sold 24 cheese pizzas and 16 specialty pizzas for a particular day. 14 of the cheese pizzas were sold to familiies with young children. 6 of the specialty pizzas were sold to familiies with young children. If a pizza was selected at random, what is the probability it was a specialty pizza or a pizza sold to a family without young children?

Select one:

a. 20/40 = 1/2

b. 26/40 = 13/20

c. 6/40 = 3/20

d. 16/40 = 2/5

Answer 1

Solution:

Given:

A pizza restaurant sold 24 cheese pizzas and

16 specialty pizzas for a particular day.

14 of the cheese pizzas were sold to families with young children.

6 of the specialty pizzas were sold to families with young children.

Thus we get following two way frequency table:

	Young Children	Without Young Children	Total
Cheese pizzas	14	=24-14 = 10	24
Specialty pizzas	6	=16- 6 = 10	16
Total	20	20	N = 40

If a pizza was selected at random, what is the probability it was a specialty pizza or a pizza sold to a family without young children?

P( specialty pizza or a pizza sold to a family without young children) =................?

Thus using addition rule:

P(specialty pizza or without young children)

=P(specialty pizza)+P(without young children)-P(specialty pizza & without young children)

=Total of specialty pizza / N + Total of without young children / N - Number specialty pizza sold to without young children / N

=16 / 40 + 20 / 40 - 10 / 40

= ( 16+20-10) / 40

= 26 / 40

= 13 / 20

Thus

P(specialty pizza or without young children) = 26 / 40 or 13/20

thus correct answer is: b. 26/40 = 13/20