Posted by **ami** on Saturday, September 29, 2012 at 6:18pm.

Suppose that you

flip a fair coin (P(H)=P(T)=1

2 ) three times and you record if it landed on heads, H, or tails,

T.

(a) What is the sample space of this experiment? What is the probability of each event?

(b) [1 pt] Let X be the number of times that you observe heads. What type of random variable is X?

(c) What is the probability distribution of X?

- Needs probability and statistics help -
**Damon**, Saturday, September 29, 2012 at 6:48pm
By the way, it is not clear to me from the wording of the question if h, t, t is the same as t, h , t or not

the first way there are 6 possible outcomes. The second way there are 8

If order does not matter

1 h h h

2 h h t

3 h t t

4 t t t

probability of each event here is 1/4

if order does matter

1 h h h

2 h h t

3 h t h

4 h t t

5 t t t

6 t t h

7 t h t

8 t h h

probability of each event here is 1/8

- stor -
**Damon**, Saturday, September 29, 2012 at 7:53pm
In the second list, listing all possible outcomes there is one way to get three heads p(3h) = 1/8

three ways of getting two heads

p(2h) = 3/8

three ways of getting 1 head

p(1h) = 3/8

one way of getting 0 heads

p (0h) = 1/8

let's look at a binomial distribution where p(h) = 1/2

n = number of trials = 3

k = number of successes (number of heads)

p(h = k) = C(n,k) p^k (1-p)^k

p(h=3) = c(3,3)(1/2)^3(1/2)^0

= 3!/[3!*0!](1/8)(1) = 1/8

p(h=1) = C(3,1)(1/2)(1/2)^2

c(3,1) = 3!/[1!(3-1)!] = 3*2/2 = 3

so

p(h=1) = 3(1/2)(1/4) = 3/8

you will find that p(h=2) = C(3,2)((1/2)^2(1/2)^1 = 3/8

etc

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