# Please help, I really need the answer to this. On a fair spinner with 8 equal sections... What is the probability of choosing 4 AND THEN choosing an odd number? (The answer must be a fraction)

You have a 1/8 probability of choosing a 4.

1, 2, 3, 4, 5, 6, 7, 8

------Independent occurrences.

You have a 1/4 chance of pulling an odd number.

1, 3, 5, 7

------

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I'd never dream of challenging John's math - and what he says is correct.

But, in a single sitting where one action occurs after the other, the odds of both happening, one after the other is 1/8 * 1/4 = 1/32

7 People thanked the writer.
John McCann commented
"AND THEN choosing"

Now I can't decide whether the emphasis (AND THEN ) mean an independent occurrence or a dependent occurrence!
Ray Dart commented
Yep, 'tis confusing.

You have two correct answers and here's why:

If A and B are independent events,
P(A and B) = P(A) • P(B).

www.regentsprep.org/regents/math/algebra/apr6/lindep.htm

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Ray Dart commented
That is the least helpful answer to that question, and the website sucks bigtime - truly awful. (Sorry, I'll shut up now)
Tom Jackson commented

John listed the probability for the two independent events.

You did the multiplication.

Both of you independently were as helpful as a sportscaster who gives a number of partial scores, to wit: 7 to 5, 21 to 17 and 2 to 1.

I listed the teams in my answer. Neither of you did in yours.

Did you click on the right URL?
Ray Dart commented