at a concession stand, the combo purchases for one week are recorded in the table below:\n|…

at a concession stand, the combo purchases for one week are recorded in the table below:\n| |hamburger|pizza|deli sandwich|\n|--|--|--|--|\n|french fries|83|67|37|\n|peanuts|2|5|14|\n|popcorn|19|29|3|\nif we choose a customer at random, what is the probability that they have purchased popcorn or a hamburger?\np(popcorn or hamburger) = \\frac{136}{?}\ngive your answer in simplest form.
Answer
Explanation:
Step1: Calculate total number of purchases
Sum all the values in the table: $(83 + 67+37)+(2 + 5+14)+(19 + 29+3)=269$
Step2: Calculate number of popcorn or hamburger purchases
Number of popcorn purchases: $19 + 29+3=51$. Number of hamburger purchases: $83 + 2+19 = 104$. But we double - counted the cases where both are purchased. The number of cases where both are purchased is $19$. So number of popcorn or hamburger purchases is $51+104 - 19=136$.
Step3: Calculate the probability
The probability $P=\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}}$. Here, the number of favorable outcomes is 136 and total number of outcomes is 269.
Answer:
$\frac{136}{269}$