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Mathematics 14 Online
OpenStudy (anonymous):

What is the probability that a randomly chosen person is female or likes McDonalds?

OpenStudy (anonymous):

OpenStudy (anonymous):

@kropot72

OpenStudy (anonymous):

Man, I was having trouble on this question, and when I saw you online, you made my day :D

OpenStudy (anonymous):

I don't know where to start on this one, its complicating. I know when its "or" I have to find the sum. But it still won't give me the answer.

OpenStudy (kropot72):

Have you completed the table by finding the sums of the rows and the columns? This is the first step.

OpenStudy (anonymous):

Yes, I know the sum of the whole table is 100, and I know that for females is 55, and that for mcdonald is 40

OpenStudy (kropot72):

Good. So we know the following: P(F) = 0.55 P(McD) = 0.4 Now you need to find the intersection of 'female' and 'likes McDonalds' on the table.

OpenStudy (anonymous):

.55/100 + .40/100 (

OpenStudy (kropot72):

So far so good. Now you need to find the intersection of 'female' and 'likes McDonalds' on the table.

OpenStudy (anonymous):

And I have to subtract them by p(female and from mcdonald) Which I think is .20/100

OpenStudy (anonymous):

Is that it, because in my book, it says the answer is 3/4, and I don't know how they came up with this answer

OpenStudy (kropot72):

The formula needed is: \[\large P(A \cup B)=P(A)+P(B)-P(A \cap B)\]

OpenStudy (anonymous):

I'm familiar with this formula, but is p(female and from mcdonald)=.20/100 right? I just want to make sure

OpenStudy (kropot72):

\[\large P(Female\ \cap McDonalds)=\frac{20}{100}=0.2\]

OpenStudy (anonymous):

Yeah, got it, I accidentally placed the decimals and it kept giving me the wrong answer.

OpenStudy (anonymous):

Thanks bud :D

OpenStudy (kropot72):

You're welcome :)

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