Josh3514idn
Answered

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A diner offers a lunch special. Customers
have a choice of a sandwich (chicken, turkey
a
veggie), a side (an apple, chips), and a drink
(water, juice). What is the probability of a
customer randomly choosing a combination
package that has chips and juice?

Sagot :

Step-by-step explanation:

First, We do the multiplication law. which states that if

R can occur multiple ways, n, then the total numbers of ways is just

[tex]s \times r[/tex]

In this case, we have 3 things to consider so we just do

number of sandwiches which is 3: Number of sides to choose from which is 2: And number of drinks: 2 so we do

[tex]3 \times 2 \times 2 = 12[/tex]

So there is 12 possible combinations. Now let see which ones contine chips and juice.

We can make a tree diagram, and see that we can do

Chicken,Chips, and Juice

Turkey, Chicken,Juice

Or Veggie,Chicken, and Juice so there is 3 combinations that includes that.

So out of 12 combinations, there is 3 that favor our scenario so we say

Probability is 3/12=1/4=25%

Mhanifaidn

Answer:

  • 1/4 or 25%

Step-by-step explanation:

The choices we are interested in are a side and a drink. The choice of a sandwich doesn't make a difference.

The probability of each is 1/2, so the probability of choosing a combination that includes chips and juice is:

  • 1/2*1/2 = 1/4 or 25%
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