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Suppose that an economist is paid to predict whether the US unemployment rate will rise or fall each time before a new government report is released. If this economists makes predictions based on the toss of a fair coin, what is the probability that they will be right exactly 9 out of the next 10 times?

I thought it was (.5)^(9)=.001953125 but it was wrong.

I'm not sure what to do. Any help would be great. Thanks!

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Ok, so the probability he is correct or wrong are both $\frac{1}{2}$ So the probability he gets any given sequence of rights and wrongs is $\frac{1}{1024}$. For example

$cwcwcwcwcw$ is one such combination.

How many different lists are there where he gets 9 correct? well the same number of lists as the ones where he gets 1 wrong. So 10.

Therefore the probability is $\frac{10}{1024}$

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  • $\begingroup$ 1/2024? Did you mean 1/1024? $\endgroup$
    – Josiah
    Feb 21 '14 at 3:43
  • $\begingroup$ yws, oops. thanks for noticing $\endgroup$
    – Yorch
    Feb 21 '14 at 3:49
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Answer:

If you want to start using formulas for such problems, here is the solution.

Probability of getting corrct p = $\frac{1}{2}$,

Probability of getting wrong q = 1-p = 1 -$\frac{1}{2}$ =$\frac{1}{2}$

Let X be the number of times he gets it correct

P(X=9) = ${10\choose9}p^9.q^{(10-9)}$ = $\frac{10}{2^{10}} = \frac{10}{1024}$

${10\choose9} = {10\choose1} = 10$ (This is what the other responder means "How many different lists are there where he gets 9 correct? well the same number of lists as the ones where he gets 1 wrong. So 10."

After understanding the problem, you shall start to use formulas like this mechanically.

Thanks Satish

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