Question 319371: What is the probability of getting between 3 and 6 heads in 10 tosses of a fair coin?
Found 2 solutions by stanbon, solver91311:
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! What is the probability of getting between 3 and 6 heads in 10 tosses of a fair coin?
------------------
P(3<= x <=6) = binomcdf(10,0.5,6)-binomcdf(10,0.5,2) = 0.7734
==========================
Cheers,
Stan H.
Answer by solver91311(24713)  (Show Source):
You can put this solution on YOUR website!
"between 3 and 6" is ambiguous. Do you mean inclusive or exclusive of 3 and 6? No matter, I'll show you what to do either way.
If you mean inclusive of 3 and 6, then the probability you seek is the probability of getting exactly 3 heads out of 10 tosses, plus the probability of getting exactly 4, plus exactly 5, plus exactly 6.
On the other hand, if you mean exclusive, then the probability you seek is the probability of exactly 4 heads out of 10 tosses plus the probability of exactly 5 heads out of 10 tosses.
The probability of  successes out of  trials where the probability of success on any individual trial is  is given by:
\ =\ \left(n\cr k\right\)p^k\left(1\,-\,p\right)^{n\,-\,k})
Where ) is the number of combinations of things taken at a time and has the value !})
For a fair coin toss, the probability of a head on any given trial is 
So, for the inclusive case, you need to calculate the following:
\ =\ \left(10\cr \,3\right\)\left(\frac{1}{2}\right)^3\left(\frac{1}{2}\right)^{7}\ =\ \left(10\cr \,3\right\)\left(\frac{1}{2}\right)^{10})
\ =\ \left(10\cr \,4\right\)\left(\frac{1}{2}\right)^{10})
\ =\ \left(10\cr \,5\right\)\left(\frac{1}{2}\right)^{10})
\ =\ \left(10\cr \,6\right\)\left(\frac{1}{2}\right)^{10})
And sum the four results.
For the exclusive case, you only need to compute
and sum the two results
John
|
|
|
