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2 votes
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

%

User Slinkp
by
7.9k points

2 Answers

9 votes

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.

User Luikore
by
8.8k points
2 votes

Answer:

4/27=14.8%

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: (2)/(3) *(2)/(3) *(1)/(3) \\Probability: (2*2*1)/(3*3*3) \\Probability: (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.

User Qwertp
by
8.2k points
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