I need to know if I use the Poisson Distribution for problem 5 (A-C).
No use the Binomial distribution p= 1/3 for tie p=1/6 for "0-0" tie
so i would use this formula: P^k * (1-p)^n-k and for part A, I would plug in 10 for n and 3 for k???
@jim_thompson5910 @karatechopper @JuanitaM
n=10 and k=3
so im correct. woohoo! :D
haha :)
ok, so i get .0021676912. is that the final answer or is there anything else i needa do?
for part A ? i get about 0.26
yeah, how'd u get .26 ??
just plugged numbers in formula ... well i cheated and used Excel :) p = 1/3 n=10 k=3 \[\rightarrow \left(\begin{matrix}10 \\ 3\end{matrix}\right)*(\frac{1}{3})^{3} *(\frac{2}{3})^{7} = 120*\frac{128}{3^{10}} = 0.2601\]
im confused about the second half... after the first equal sign
\[\left(\begin{matrix}10 \\ 3\end{matrix}\right) = \frac{10!}{7! 3!} = 120\] \[(\frac{1}{3})^{3} *(\frac{2}{3})^{7} = \frac{1}{3^{3}}*\frac{2^{7}}{3^{7}} = \frac{2^{7}}{3^{10}} = \frac{128}{59,049}\]
ok. so i was getting .0021676912 because i didn't multiply it by 120. makes sense.
so for part B i would just change p to 1/6 ??? (why 1/6 again?) and for part C i would do 1-part B ???
its 1/6 because its half of 1/3 --> 1/3 * 1/2 = 1/6 part C is looking at chance you tie but Not 0-0 --> 1/3 - 1/6 = 1/6
o because half would equal zero-zero ??
right
and part Cs answer will basically be the same as part B since they're both p=1/6?
umm yep thats what it looks like
ok coo ! :)
alright, so im on #7 and it's wanting the same info as #6 but normally distributed... would i use : f(x)= ( 1 / [sigma * sqrt(2pi)] ) * e ^ (x-mu/sigma) to plug in the same info from 6???
here are the answers to #6: a) picture b) 1/(b-a) = 1/(210-150) = 1/60 c) (160-150)/(210-150) = 10/60 = 1/6 d) 1/6 * 1/6 = 1/36 e) (185-165)/(210-150) = 20/60 = 2/6 f) (210-200)/(210-150) = 10/60 = 1/6
|dw:1360115375877:dw|
Join our real-time social learning platform and learn together with your friends!