Answer:
In the long run, ou expect to lose $4 per game
Step-by-step explanation:
Suppose we play the following game based on tosses of a fair coin. You pay me $10, and I agree to pay you $n^2 if heads comes up first on the nth toss.
Assuming X be the toss on which the first head appears.
then the geometric distribution of X is:
X
geom(p = 1/2)
the probability function P can be computed as:

where
n = 1,2,3 ...
If I agree to pay you $n^2 if heads comes up first on the nth toss.
this implies that , you need to be paid 

![\sum \limits ^{n}_{i=1} n^2 P(X=n) =Var (X) + [E(X)]^2](https://tex.z-dn.net/?f=%5Csum%20%5Climits%20%5E%7Bn%7D_%7Bi%3D1%7D%20n%5E2%20P%28X%3Dn%29%20%3DVar%20%28X%29%20%2B%20%5BE%28X%29%5D%5E2)
∵ X
geom(p = 1/2)








Given that during the game play, You pay me $10 , the calculated expected loss = $10 - $6
= $4
∴
In the long run, you expect to lose $4 per game
Answer:
Step-by-step explanation:
Comment
Let the number of pages = x
Play = 1/3x
Actors = 1/4(x - 1/3 x)
Actors = 1/4(2/3x)
Actors = 2/12 x
Producer Director etc = 2 pages
Equation
Actors = 2 pages
2/12 x = 2
Solution
2x/12 = 2 Multiply both sides by 12
12*2x/12 = 2*12 Combine
2x = 24 Divide by 2
2x/2 = 24/2 Combine
x = 12
Answer
The program had 12 pages.
Answer:
Step-by-step explanation:
Total fruits T= 5+7 = 12
No. Of apple n(A) = 5
No. Of peaches n(P) = 7
P(A) 1st = n(A)/T = 5/12
P(P) 2nd = n(p)/(T-1) = 7/11
: P(A and P) = 5/12 × 7/11
= 35/132
Use the quadratic formula -b+/-√b^2 -4(ac) / 2a
-4 +/- √(4^2 - 4(1*10) / (2*1)
X = -2 +/- i√6