Please Help Stats Hw?

Question

How do you solve this problem? using Permutation? or Combination...

A coin is tossed 10 times and the resulting sequence of heads and/or tails is recorded. how many sequence have:

a. exactly 4 heads?
b. exactly 7 heads?
c. at least 8 heads?
d. at least 2 heads?

Answer 1

You use Combinations. Not sure if you are using a calculator for these, but the answers are:

a) C(10,4) b) C(10,7)
c) C(10,8) + C(10,9) + C(10,10)
d) C(10,2) + C(10,3) + C(10,4) + . . . . .+ C(10,10)

Hope this helped!]

Answer 2

Use the poisson distribution with p=0.5, q=0.5
where p is the probability of tossing a head.
Since all combinations of pq= 0.5^10, 1/1024, this can be factored out of the distribution to give:
(1/1024) (1+10+45+120+210+252+........1)
p(4 heads)= 210/1024
p (7 heads=120/1024
p(8,9,10 heads)=56/1024
p(0 to 8 tails)=968/1024

Answer 3

D..

Its all odds. Well, a coin is two sided and you can not tell exactly what it will be. So the odds are 50/50 and you have to go by that. You can't tell if it is exact, so that eliminates a and b. You know it can't be 8 because its over 5/10... which leaves d.

I hope that helps!

Answer 4

i'm over forty, nonetheless a Democrat yet, the Democratic occasion has moved too far left even for me so my _label_ is now Republican (which occasion's platform is now only approximately same to the Democratic platform and philosophy of the 1960's).

Answer 5

Here's a different way to go about it.

Use the binomial and multiply by the total number of outcomes, which 2^10.

a): [C(10,4)(0.50)^4(0.50)^6]*2^10=210
b): [C(10,7)(0.50)^7(0.50)^3]*2^10=120
c): [SUM(10,k)(0.50)^k*(0.50)^(10-k), k=8..10]*2^10=56
d): [SUM(10,k)(0.50)^k*(0.50)^(10-k),k=2..10]*2^10=1013