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Intermediate Probability

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Intermediate Probability

by Aman verma » Sun Mar 21, 2010 3:33 am
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
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by rohan_vus » Sun Mar 21, 2010 4:54 am
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]
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by Stuart@KaplanGMAT » Sun Mar 21, 2010 1:32 pm
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)!
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by Aman verma » Mon Mar 22, 2010 2:01 am
OA D. Thanks to all !
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by GMATSUCKER » Mon Mar 22, 2010 3:04 am
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 !!!!!!!!!!
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by Haldiram Bhujiawala » Mon Mar 22, 2010 3:49 am
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 !!!
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by mj41 » Tue Mar 23, 2010 9:31 am
rohan_vus can you please explain the equation you used
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