Patrick B. answered 01/31/21
Math and computer tutor/teacher
a) Then X =0,1,2,3
b)
Prob(X=0) = 1/2 * 1/2*1/2 = 1/8
the head can appear in 1 of 3 positions
Prob(X=1) = 1/2 * 1/2*1/2 *3 = 3/8
the tail can appear in 1 of 3 positions
Prob(X=2) = 1/2*1/2*1/2 *3 = 3/8
Prob(X=3) = 1/2*1/2*1/2 = 1/8
c)
0 *
1 ***
2 ***
3 *
d)
expected value
= SUM OF [ outcome * prob]
= 0*1/8 + 1*3/8 + 2*3/8 + 3*1/8
= 3/8+6/8+3/8
= 12/8
= 3/2
e)
Var = Expected Value of x^2 - (Expected value of x)^2
Expected value of x^2 =
= 0^2 * 1/8 + 1^2 * 3/8 + 2^2 * 3/8 + 3^2*1/8
= 1*3/8 + 4*3/8 + 9*1/8
= 3/8 + 12/8 + 9/8
= 24/8
= 3
variance = 3 - 3/2 ^2
= 3 - 9/4
= 12/4 - 9/4
= 3/4
f) standard deviation is the square root of the variance
sqrt(3/4) = sqrt(3)/2
g) Prob(X>=2) = 3/8+1/8 = 4/8 = 1/2 = 50%
so h) No it is likely to happen