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September 18, 2014

September 18, 2014

Posted by **Amelie** on Sunday, February 17, 2008 at 3:35pm.

(a) List the sample space.

(b) Determine the probability function of X.

======================

Answer: a) 2^3= 8 possibilities.

b) (P(heads 1st time)+ P(heads 2nd time) +P( heads 3rd time)) / 15

since each probability has 1/2 chance of occuring, the answer is 1/10

- Intro to Probability, please check my work. -
**drwls**, Sunday, February 17, 2008 at 3:54pma) is correct

b) the Probability function is:

1/8: zero heads, $0 won

3/8: one heads, $5 won

3/8: two heads, $10 won

1/8: three heads, $15 won

Average winnings: 30/8 = $7.50

- Intro to Probability, please check my work. -
**Amelie**, Sunday, February 17, 2008 at 3:57pmthank you drwls

- Intro to Probability, please check my work. -
**Damon**, Sunday, February 17, 2008 at 4:38pmSpace:

t t t

t t h

t h t

t h h

h t t

h t h

h h t

h h h

So there are 8 possible patterns, agree with 2^3

there is One way to get 0 heads so 1/8 at 0

there are three ways to get 1 heads so 3/8 at 5

also

3/8 at 10

and

1/8 at 15

I get average = (1/8)(1*0+3*5+3*10+1*15) = 60/8

By the way, this is a binomial distribution problem

The probability of k heads in n tosses =

C(n,k)p(heads in a toss or .5))(1-p) which is also .5 for our coins

here C(n,k) = n!/[k!(n-k)!]

for n = 3 tosses

3! for example = 3*2*1 = 6

0! = 1 by convention

C(3,0) = 3!/[0!*3!) = 1

.5^0*.5^3 = 1/8

so p(0 heads) = 1*1/8 = 1/8 as we knew

C(3,1) = 6/[1!(2!)] = 6/2 = 3

.5^1*.5^2 = .125 = 1/8

so p(1 heads) = 3/8 as we knew

etc

now many of us know those coefficients by heart for small n, but you can also get the from making "Pascal's triangle

1 row zero

1 1 row 1

1 2 1 row 2

1 3 3 1 row 3 the one we want !

the zero and right elements are 1

between, each element is the sum of the two above it right and left.

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