Question

Edna plans to treat her boyfriend curt to dinner for his birthday. the costs of their date options are listed next to each possible choice. edna plans to allow curt to choose whether they will eat mexican food ($25), chinese food ($15), or italian food ($30). next, they will go bowling ($20), go to the movies ($30) or go to a museum ($10). edna also is deciding between a new wallet ($12) and a cell phone case ($20) as possible gift options for curt. what is the probability that edna spends more than $55 on the date?

Answer 1

There is a 61.11% chance that Edna will spend more than $55 on this date.

There are three options for food, three options for entertainment and two choices for gifts.

Hence, the total number of decisions in this case is 3 *3 *2 = 18 decisions.

The costs associated with each of these options is given in the attachment below.

Out of a total of 18 decisions, the cost of 11 decisions exceed $55.

$Probability of an event = \frac{No. of events favouring the event}{Total number of events}$

$P(Cost > 55) = \frac{11}{18}$

$P(Cost > 55) = 0.611111111$