Probability of Head for each coin = P
Probability of not having Head for each coin = (1-P)

Probability of 50 Heads = (p)^50*(1-P)^50*100!/(50!*50!) ---- (1)
Probability of 51 Heads = (P)^51*(1-P)^49*100!/(51!*49!) ----(2)

Per question st1 and St2 are same , so (p)^50*(1-P)^50*100!/(50!*50!) = (P)^51*(1-P)^49*100!/(51!*49!)
==> (1-P)/50 = P/51
==> 51 - 51P = 50P
[spoiler]so P = 51/101
D is the answer[/spoiler]

Aman verma wrote:Q: One hundred identical coins each with probability P of showing up Heads are tossed once. If 0<P<1 and the probability of Heads showing on 50 coins is equal to that of Heads showing on 51 coins, then the value of P is :

a)1/21

b)49/101

c)50/101

d)51/101

e)61/97

Without doing any math:

If there was a 1/2 chance of getting heads, then the most frequent number of heads should be 50 out of 100.

In this question, the chance of getting 50 heads is the same as the chance of getting 51 heads; therefore, the coin must be slightly weighted toward heads.

So, we want an answer choice a tiny bit higher than 1/2: only (d) is in the ballpark, choose (d)!

OA D. Thanks to all !

Stuart Kovinsky wrote:
Without doing any math:

In this question, the chance of getting 50 heads is the same as the chance of getting 51 heads; therefore, the coin must be slightly weighted toward heads.

So, we want an answer choice a tiny bit higher than 1/2: only (d) is in the ballpark, choose (d)!

Love that approach !!!!!!!!!!

GMATSUCKER wrote:

Stuart Kovinsky wrote:
Without doing any math:

In this question, the chance of getting 50 heads is the same as the chance of getting 51 heads; therefore, the coin must be slightly weighted toward heads.

So, we want an answer choice a tiny bit higher than 1/2: only (d) is in the ballpark, choose (d)!

Love that approach !!!!!!!!!!

Me Too !!!

rohan_vus can you please explain the equation you used