Read about the binomial probability distribution.
-- Rick
--- Sabrina <gummiwulf@yahoo.com> wrote:
> Thank you Brian! I love all of the details in your response. It's
> great to see so many people that love math. :) I just wish that I
> had a way of wrapping my mind around probability... Let me see if I
> understand your answer.
>
> So let's say I rolled this die for a sets of 6 rolls. As the die
> has a 6% chance each roll of landing on value C, each set of 6 rolls
> that I create has a .257% chance of having exactly 3 Cs in the
> outcome. Is that right? Or is it 2.57%?... Sheesh I need to go
> back to reading more probability!
>
>
>
> --- In Math4u@yahoogroups.com, "Brian Edward Jensen"
> <brianejensen@...> wrote:
> >
> > Suppose you are like me and know nothing about probability.
> >
> > One way to solve this would be to throw the die 1.0 gazillion sets
> of
> > 6. This would take a long time. Let's call this certain
> number "c."
> > After the first throw, we will have
> >
> > One c (3/8) gazillion times
> >
> > Zero c (5/8) gazillion times.
> >
> > After two throws, we could have
> >
> > Two c's (3/8)^2 gazillion times
> >
> > One c [(3/8)*(5/8)+(5/8)*(3/8)] gazillion times. = 2(3/8)(5/8)
> > gazillion times
> >
> > Zero c's (5/8)^2 gazillion times.
> >
> > Discovery, this looks like Pascal's triangle. Keep doing this
> until
> > you see the pattern.
> >
> > 1 toss: 1,1
> >
> > 2 toss: 1,2,1
> >
> > 3 toss: 1,3,3,1
> >
> > 4 toss: 1,4,6,4,1
> >
> > 5 toss: 1,5,10,10,5,1
> >
> > 6 toss: 1,6,15,20.15.6.1
> >
> > In the past, I have done this on a spreadsheet. It looks to me
> like
> > after 6 tosses,
> >
> > The odds of getting 6 c's is 1*[(3/8)^6] The first number is from
> > Pascal's triangle.
> >
> > The odds of getting 5 c's is 6*[(3/8)^5*(5/8)]
> >
> > The odds of getting 4 c's is 15*[(3/8)^4*(5/8)^2]
> >
> > The odds of getting 3 c's is 20*[(3/8)^3*(5/8)^3] ANSWER
> >
> > The odds of getting 2 c's is 15*[(3/8)^2*(5/8)^4]
> >
> > The odds of getting 1 c's is 6*[(3/8)^1*(5/8)^5]
> >
> > The odds of getting 0 c's is 1*[(3/8)^0*(5/8)^6]
> >
> > Check
> >
> > The odds of getting 6 c's is 0.002781
> >
> > The odds of getting 5 c's is 0.027809
> >
> > The odds of getting 4 c's is 0.115871
> >
> > The odds of getting 3 c's is 0.257492 ANSWER
> >
> > The odds of getting 2 c's is 0.321865
> >
> > The odds of getting 1 c's is 0.214577
> >
> > The odds of getting 0 c's is 0.059605
> >
> > Total odds is ++++++++++1.0
> >
> > I think it is amazing that such equations will add up to 1.0
> >
> > Another way
> >
> > The odds that the first 3 throws will get the desired number and
> the
> > next 3 won't get the desired number is
> >
> > (3/8)^3*(5/8)^3
> >
> > Now I interpreted the problem to say we want the correct number to
> > come up EXACTLY 3 times. If they mean at least 3 times, then we
> need
> > to add the odds of getting the correct number 3, 4, 5, or 6 times
> > above.
> >
> > We need to multiply this number times the different ways of
> picking 3
> > correct numbers. The first correct number has 6 throws where it
> can
> > occur. The next one has 5 throws where it can occur. The last has
> 4
> > throws where it can occur. So we have 6*5*4. But some of these
> > combinations are indistinguishable from each other. These 3
> correct
> > numbers can be interchanged 3*2 different ways. So we actually
> have
> > 6*5*4/(3*2) different ways of getting 3 correct numbers with 6
> throws
> > of the die. This equals 20 which is the same number we picked out
> of
> > Pascal's triangle with the first method!
> >
> > So the answer is
> >
> > 6*5*4/(3*2)*[(3/8)^3*(5/8)^3]
> >
> > =20*[(3/8)^3*(5/8)^3]
> >
> > =0.257492 ANSWER
> >
> > Now if I compare this to the "choose and pick" language that
> others
> > will write in, I will understand it even better.
> >
> > Regards,
> >
> > Brian
> >
> >
> > --- In Math4u@yahoogroups.com, "Sabrina" <gummiwulf@> wrote:
> > >
> > > Hello everyone. :) I was wondering if I could get some
> assistance
> > on
> > > how to calculate the following:
> > >
> > > Say I have a 3/50 chance of rolling a certain number on a die
> (or
> > > generating a number in a random number generator, etc). I roll
> the
> > > die 6 times, and come up with that number 3 times. How do I
> figure
> > > out what the odds are of hitting that number 3 times in 6 rolls?
> > >
> > > Thank you so much!
> > > ~gummiwulf
> > >
> >
>
>
>
>
>
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