could somebody do question b
b) Probability of 3 or more blue in 4 trials.Originally Posted by Arbitur
Probability of a blue on any trial is 0.8 as we are sampling with replacement,
so:
$\displaystyle
P(N_{blue} \ge 3)=P(N_{blue} = 3)+P(N_{blue} = 4)
$
but:
$\displaystyle
P(N_{blue} = K\ \mbox{in M trials})={M \choose K}p^K(1-p)^{M-K}=$$\displaystyle \frac{M!}{K!(M-K)!}p^K(1-p)^{M-K}
$
Here $\displaystyle p=0.8$, so:
$\displaystyle
P(N_{blue} \ge 3)=\frac{4!}{3!1!}\ 0.8^3\ 0.2^1+\frac{4!}{4!0!}\ 0.8^4\ 0.2^0
$
Which since $\displaystyle 1!=0!=1$ may be reduced to:
$\displaystyle
P(N_{blue} \ge 3)=4\ 0.8^3\ 0.2+0.8^4
$
RonL
uhm can you explain that in words?
i need to get it to a percent number
i thought it had to be something like
mean=.8*4-3.2
standard deviation=square root of 4*.8*.2
=.64
so (SOME NUMBER)-3.2/.64 =the zscore i need to get the percentage???
that is what my book is teaching me
what number do i use
Then you are probably looking at the wrong section of your book. YouOriginally Posted by Arbitur
should be looking at the section that deals with the Binomial distribution.
See the last line of my post:what number do i use
$\displaystyle
P(N_{blue} \ge 3)=4\ 0.8^3\ 0.2+0.8^4
$
That can be simplified to the number you want:
$\displaystyle
P(N_{blue} \ge 3)=4\ 0.8^3\ 0.2+0.8^4\approx 0.819
$
RonL
thanks anyways but i have no idea how u get that ;/
ok im looking through my book and
because of a rule i cant use the z-score way and crap... but how do i know what way and numbers to use then
ok i get it now... but what if b used really big numbers... would it still work that easy? if not what would i do( cause those are easily added 2 gether)
With really "big numbers" you would use the normal approximation to theOriginally Posted by Arbitur
binomial distribution; see this thread
RonL
That the normal approximation is not too good for small problem sizes canOriginally Posted by CaptainBlack
be shown by reworking this problem using the normal approximation.
Here $\displaystyle p=0.8$, and $\displaystyle N=4$, so the mean number
of blues is:
$\displaystyle
\mu_B=0.8 \times 4=3.2
$
The vaiance of the number of blues is:
$\displaystyle
\sigma_B^2=4\times 0.8 \times 0.2=0.64
$
Now we want the probability of $\displaystyle 3$ or more blues, so we want
the critical z-score for $\displaystyle 2.5$ (rounding down by $\displaystyle 0.5$
as a continuity correction):
$\displaystyle
z_{critical}=\frac{2.5-3.2}{0.8}=-0.875
$,
Look this up in a normal table we find:
$\displaystyle
P(z>-0.875)\approx 0.809
$
rather than the correct value of $\displaystyle \sim 0.819$
RonL