A fair coin is to be tossed 100 times with each toss resulting in a head or a tail. If H is the total number of heads and T is the total number of tails, which of the following events has the greats possibility?

If the probability of success is "p" and the probabibility of failure is "q", then the probability of at least "k" successes on n trials is C(n, k) *
p^k * q^(n - k)

Using the above we find n = 100, p = q = 1/2 = 0.5 and,

A. H = 50,
P = C(100, 50) * (1/2)^50*(1/2)^(100-50) = C(100, 50)/2^100

B. T >= 60, P = C(100, 60) * (1/2)^40*(1/2)^60 + C(100, 61) * (1/2)^39 * (1/2)^ 61 + ...
P = [ C(100, 60) + C(100, 61) + C(100, 62) + .... + C(100, 100) ]/2^100

C. 51 <= H <= 55,
P = [ C(100, 51) + ... + C(100, 55) ] / 2^100

D. H >= 48 and T >=48
P = [ C(100, 48) + C(100, 49) + C(100, 50) + C(100, 51) + C(100, 52) ] / 2^100

E. H <=5 or H >=95
P = [ C(100, 0) + C(100, 1) + ... + C(100, 5) + C(100, 95) + ... + C(100, 100) ] / 2^100

Now using the properties of C(n, k) = n!/[k!(n - k)!], the above may be somewhat simplified to,

A. P = C(100, 50)/2^100

B. P = [ C(100, 60) + C(100, 61) + C(100, 62) + .... + C(100, 100) ]/2^100
= 1 - [ C(100, 0) + ... C(100, 49) + C(100, 50) + C(100, 51) + ... + C(100, 59) ] / 2^100

C. P = [ C(100, 51) + ... + C(100, 55) ] / 2^100

D. P = [ C(100, 48) + C(100, 49) + C(100, 50) + C(100, 51) + C(100, 52) ] / 2^100

E. P = 2 [ C(100, 0) + C(100, 1) + ... C(100, 5) ] / 2^100

Which one is the greatest? Well to be precise use a calculator. Or...
Observe that C > A, D > A, E can easily be calculated to be small, B although large is far away from the middle value C(100, 50).

Elementarily, I know that the combinatorial function takes its greatest value at the midpoint.
E.g. C(2, 0) = 1, C(2, 1) = 2, C(2, 2) = 1
C(7, 0) = 1, C(7, 1) = 7, C(7, 2) = 21, C(7, 3) = 35, C(7, 4) = 35, ....

So I would bet on D. as it takes the largest number of values from the middle of the range for C(100, k).