Question

Lunches can be made consisting of exactly one entrée, one drink, and one snack. If all lunch options are equally likely, what is the probability of randomly choosing a lunch that
does not contain the veggie wrap
has milk or water, and
has fruit as the snack.
Write your answer as a percentage, to the nearest percent.

Entrée Chicken Cutlet Veggies Wrap Lasagna
Drink Milk Water Iced Tea
Snack Fruit Chips Jello

%

Answer 1

Answer:

4/27=14.8%

Step-by-step explanation:

In order to solve this, we have to see first each option possible, there are 2/3 of possibilities that a random Entrée ends up being Chicken Cutlet or Lasagna, then you have again 2/3 chances to get milk or water, and as you only want fruit the chances are 1/3 that you get that.

In order to calculate if this could happen at random we have to multiply all of the odds:

$Probability: \frac{2}{3} *\frac{2}{3} *\frac{1}{3} \\Probability: \frac{2*2*1}{3*3*3} \\Probability: \frac{4}{27}$

So we know that for every 27 times you pick a lunch at random, 4 times it will not have the Veggie Wrap, contain milk or water, and will have fruit as the snack.

Answer 2

There are 2/3 of possibilities that a random Entrée ends up being Chicken Cutlet or Lasagna, then you have again 2/3 chances to get milk or water, and as you only want fruit the chances are 1/3 that you get that.

We need to now multiply the odds:

2/3x2/3x1/3

2x2x1/3x3x3

4/27

So we know that for every 27 times you pick a lunch at random, 4 times it will not have the Veggie Wrap, contain milk or water, and will have fruit as the snack.

So now we know it is, %15 rounded but 14.8% not rounded.