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If 20 customers are eating at an restaurant 14 peoples order pizza (P(A)), 10 peoples order salad (P(B)), 6 peoples order pizza and salad (P(A and B))

P(A) = 14/20 = 7/10

P(B) =10/20 = 1/2

P(A and B) = 6/20 = 3/10

P(A or B) = 14/20 + 10/20 - 6/20 = 18/20 = 9/10

P(neither A nor B)= 1-P(A or B) = 1/10

But I can't figure it out who neither order pizza nor salad! Every one have order something at least one.

Does P(not A or not B) exist????

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    What does it mean for a probability to exist? Do you mean "is it non-zero"? – 2017-02-20
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    Yes !! It is my miscalculation. I have confused that all of customer have order so I mean 1-P(A or B) = 0 but It's my wrong result Thank you! – 2017-02-21

2 Answers 2

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To add to Arthur's answer.

Your statement which says,

Every one have order something at least one.

is untrue. Since $14$ people ordered pizza, out of these set of people $6$ have ordered salad also. $4$ people have only salad (dietitians).

So remaining $2$ are having nothing. ($2/20$, in your expressions this is "P(neither A nor B)")

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    Thank you so much!!! I got it Thank you!! – 2017-02-21
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Hint: "not A or not B" is equivalent to "not (A and B)", the same way you've already used that "not A and not B" ("neither A nor B") is equivalent to "not (A or B)" in your last line of calculations.

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    Oh thank you so much!!! – 2017-02-21