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trasher [3.6K]
2 years ago
14

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

%
Mathematics
2 answers:
Licemer1 [7]2 years ago
8 0

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.

valentina_108 [34]2 years ago
7 0

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.

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