Is it possible that P(A) = 17/3? Why or why not?

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Is it possible that P(A) = 17/3? Why or why not?

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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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Are we talking about probability? Remember probability can only be in between 0 to 1. 0 means it's certain that it won't happen and 1 means it's certain that it will happens.
Is 17/3 in between 0 and 1?
I don't know, this is a really confusing subject for me.

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P(A) can only be in between 0 and 1, right? 0.5, 0, 1, 0.3324, or 0.65, etc. Can P(A) be equal to 17/3 ?
I would suggest you to convert 17/3 to decimal.
|dw:1436583990368:dw|
5.666667 this would be decimal form of it
Yeah. And is that number in between 0 and 1?
No, so it would be it because it is greater than 1, correct?
That's right. Therefore \(P(A)=17/3\) is invalid. Is that clear?
Yes, it is. Thanks!

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