calculate the probability of tossing a coin 15 times and getting the given number of heads. round the answer to 4 places.
please advise what does that mean.
what is the given number of heads?
the answer is the binomial im sure
the number of heads are 4, 7 and 10
15 C #heads * .5 right?
15.14.13.12 ----------- * 1/2 4.3.2 15.14.13.12.11.10.9 ------------------- * 1/2 7.6.5.4.3.2 15.14.13.12.11 --------------- * 1/2 5.4.3.2
are you doing factorial? please advise
i am.. 15 P 4 ------ * 1/2 4! 15 P 7 ------ * 1/2 7! 15 P 5 ------ * 1/2 5!
\[_{15}C_{n}(.5)^{15}\] where n is the number of heads
i dont think its .5^(15) tho. that looks odd to me
or is it nCr.p.q?
I aint got me ti83 handy to check :)
\[P(X=x)=_{n}C_{x}p^x(1-p)^{n-x}\] follows the binomial distribution
i have a ti83 to check....how do you put that information in?
2nd, vars, binomialpdf(15,.5), sto, 2nd, 2 then fill in list 1 with 0,1,2,3,4,5, ... , 14,15 then stats, calc, 1-vars, L1,L2
that last step is prolly off ... L1 tells you the number of head; and the binomial of L2 gives the probability
1-variables stat is usually to get the mean and sd and such ...
Let n=15 and x= # of heads \[P(X=x)=_{n}C_{x}p^x(1-p)^{n-x}\] \[=_{15}C_{x}(.5)^x(1-.5)^{15-x}\] \[=_{15}C_{x}(.5)^x(.5)^{15-x}\] \[=_{15}C_{x}(.5)^{x+15-x}\] \[=_{15}C_{x}(.5)^{15}\]
when do press enter?
\[_{15}C_{x}=\frac{15!}{x!(15-x)!}\]
that way is good too :) binomialpdf(15,.5,#heads) is the direct answer without the list
binomialpdf(15,.5,4), enter for 4 heads binomialpdf(15,.5,7), enter for 7 heads binomialpdf(15,.5,10), enter for 10 heads
2nd, enter gets you back to what you did prior without having to go thru the distr menu again
good answers to you both
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