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Mathematics 15 Online
OpenStudy (anonymous):

okay last time I'm going to ask this...a die is rolled 5 times. find the probability p(of exactly 3 occurrences of 6) a. 1.86% b. 2.08% c. 2.23% d. 3.22%

OpenStudy (anonymous):

Do you know the formula for Bernoulli trials? http://en.wikipedia.org/wiki/Bernoulli_trial

OpenStudy (anonymous):

no:(...i've tried to make sense of it but its just not processing

OpenStudy (anonymous):

The event of interest is rolling a 6 What is the probability (in a single roll)

OpenStudy (anonymous):

The probability of rolling a 6 is 1/6 (six-sided dice) Call that p The probabiliity of not rolling a 6 is (1-p) What do you think n and k stand for in the formula?

OpenStudy (amistre64):

is bernie the same as a binomial distribution?

OpenStudy (anonymous):

i think formulas are overrated

OpenStudy (amistre64):

formula one racers would disagree :)

OpenStudy (anonymous):

the probability you roll a six on one die is \(\frac{1}{6}\) and the probability you do not roll a six is therefore \(\frac{5}{6}\)

OpenStudy (anonymous):

but maybe not nascar drivers

OpenStudy (amistre64):

they got nascar in the olympics yet?

OpenStudy (anonymous):

@amistre64 Yes

OpenStudy (anonymous):

That's the way they index it in Wikipedia

OpenStudy (amistre64):

.... yes bernie is binomial; yes formula one racers disagree; or, yes nascar is in the plympics? :)

OpenStudy (anonymous):

multiple choice! A

OpenStudy (anonymous):

in any event, since the rolling of a die many times is an example of independent events (the dice have no memory) then what do you want? 3 sixes and 2 not sixes since the events are independent, you multiply the probabilities together to get \[(\frac{1}{6})^3\times (\frac{5}{6})^2\] which would give you the probability of rolling 3 sixes in a row and ten two not sixes but you do not need them to be in a row, you just need 3 sixes and 2 not sixes the number of way to choose 2 out of 5 (or 3 out of 5) is \(\dbinom{5}{2}=\frac{5\times 4}{2}=10\)

OpenStudy (anonymous):

@satellite73 Thanks for working through it for him.

OpenStudy (anonymous):

I just get a little discouraged when there is not any help from the OP.

OpenStudy (anonymous):

and so your answer would be \[20\times \left(\frac{1}{6}\right)^3\times \left(\frac{5}{6}\right)^2\] which is what the formula tells you, but also what you get by thinking

OpenStudy (anonymous):

thanks:)

OpenStudy (anonymous):

then a calculator would be helpful hope all the steps are clear, and more importantly that the formula \[P(x=k)=\dbinom{n}{k}p^k(1-p)^{n-k}\] makes some sense do not get married to the formula, make sure to understand what it really says

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