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Mathematics 17 Online
OpenStudy (anonymous):

a poll shows that 76% of voters in a city favor an initiative to increase spending on public schools. if 10 voters are selected at random, wht is the probability that exactly five of them will vote in favor of the initiative?

OpenStudy (anonymous):

binomial for this one

OpenStudy (anonymous):

5 will be in favor , 5 will not \[P(X=5)=\binom{10}{5}(.76)^5\times (.24)^5\] and a calculator

OpenStudy (anonymous):

or wolfram http://www.wolframalpha.com/input/?i=P%28X%3D5%29%3D \binom{10}{5}%28.76%29^5\times+%28.24%29^5

OpenStudy (anonymous):

im lost.....

OpenStudy (anonymous):

okay lets go slow

OpenStudy (anonymous):

ok

OpenStudy (anonymous):

"a poll shows that 76% of voters in a city favor an initiative " says the probability that a person is in favor of the initiative is \(.76\)

OpenStudy (anonymous):

and therefore the probability they are not is \(1-.76=.24\) is that part okay?

OpenStudy (anonymous):

yes

OpenStudy (anonymous):

ok good, now do you know how to compute a "binomial" probability? i am guessing "no"

OpenStudy (anonymous):

no

OpenStudy (anonymous):

you have 10 people all together you want to know the probability that exactly 5 are in favor, which of course means the other five are not

OpenStudy (anonymous):

ok

OpenStudy (anonymous):

you multiply all the probabilities together that is, you take \((.76)^5\) and multiply it by \((.26)^5\) that is for the five that are in favor, and the five that are not

OpenStudy (anonymous):

3.01

OpenStudy (anonymous):

that would be something like having the first 5 in favor, and the second 5 not in favor they would vote (yes, yes, yes, yes, yes, no, no, no , no , no)

OpenStudy (anonymous):

but there are other possible arrangement of this, for example (yes, no, yes, yes, no, yes, no, no , no , no, yes)

OpenStudy (anonymous):

the question is "how many" and that is the same as asking 'out of 10 people, how many ways are there to choose 5?" which is sometimes written as \(\binom{10}{5}\) and sometimes written as \(_{10}C_5\) have you seen either of these?

OpenStudy (anonymous):

the second one

OpenStudy (anonymous):

okay then we will use that

OpenStudy (anonymous):

typo there, your probability is therefore \[_{10}C_5\times (.76)^5\times (.24)^5\]

OpenStudy (anonymous):

ok but im not sure how to finish it with the first part.

OpenStudy (anonymous):

you will need a calculator for this \(_{10}C_5=525\) so on your calculator compute \[252\times (.76)^5\times (.24)^5\]

OpenStudy (anonymous):

sorry, i mean \(_{10}C_5=252\)

OpenStudy (anonymous):

http://www.wolframalpha.com/input/?i=10+choose+5

OpenStudy (anonymous):

ok if i take and times them i do not come up with any of the multiply choice answers

OpenStudy (anonymous):

here is your "final answer" http://www.wolframalpha.com/input/?i=%2810+choose+5%29+*%28.76%29^5*%28.24%29^5

OpenStudy (anonymous):

answer as you can see is about \(.0509\)

OpenStudy (anonymous):

would it be 5.1% then

OpenStudy (anonymous):

as a percent, yes

OpenStudy (anonymous):

i got it and thank you

OpenStudy (anonymous):

yw

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