CareerCup

I disagree with the answer posted above by "Anonymous"

Problem space: Person A has 100 possible change amounts - 0 through 99. The same applies for person B.
Thus possible outcomes = 100 * 100 = 10000

Favorable outcomes have to include having full dollar amounts hence (0,0) has to be a possible solution apart from the other pairings making the favorable outcome space - {(0,0); (1, 99); (2,98);... (98,2), (99,1)}

Probability = 100/10000 = 1/100

This is similar to two dice having a sum prob.... Total no. of cases are 99*99 and possible no of cases = 99 ((1,99),(2,98)...,(98,2),(99,1)). So prob will be 1/99

the sum of change could be anything 0 to 99 = 100 outcomes....one favorable outcome = 0.
so, probability=1/100

1/100

No ware in question it says X = Y = 100, so all answers are based on the assumptions that X = Y= 100...
So to be in terms on X Y its P(X)U p(Y) = P(X U Y) and actual answers depends on valus of X and Y for 100 what is mentioned above is right.

ans should be

(2* 98 + 2)/ (100^100)

Let's make a matrix. We are looking for X+Y=100 or 0. Clearly there are 100 such entries.

my mistake..
but still i m struggling
i guess the ans should be 2/100

since one case be when X+Y=0 (e.g X= 3.0 & Y=9.0) and the other one is obvious when X+Y=100 (X=9.50 & Y=7.50)

Do we care how much each spent? I would say 1/100.

correct: right answer is 1/100

2/100

100^2 possible combinations, 200 "good" ones.

consider two sets {1,2,....99} and {0}
A's expenses could be either an integer i.e.{0} or decimal number {1,2,...99}
Case 1
A's expenses are integer
i.e. B's expenses should also be integer
1/2x1/2

Case 2
A's expenses is a decimal number
then b's expenses should be the decimal inverse to make the sum zero(100).
1/2x1/99

either case 1 occurs or case 2

(1/2x1/2)+(1/2x1/99)

A = 0 ~ 99
B = 0 ~ 99

P(A + B = 0) = 1 / 10000

For any A, we can find a B where A+B = 100, i.e (A=1,B=99) or (A=2,B=98) ...
P(A + B = 100) = 100 / 10000 = 1/100

So P(0 cents) = 1.01%