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
Intermediate Probability
This topic has expert replies
-
- Master | Next Rank: 500 Posts
- Posts: 304
- Joined: Wed Jan 27, 2010 8:35 am
- Location: International Space Station
- Thanked: 11 times
- Followed by:3 members
-
- Master | Next Rank: 500 Posts
- Posts: 189
- Joined: Thu Apr 03, 2008 2:03 pm
- Location: USA
- Thanked: 21 times
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]
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]
- Stuart@KaplanGMAT
- GMAT Instructor
- Posts: 3225
- Joined: Tue Jan 08, 2008 2:40 pm
- Location: Toronto
- Thanked: 1710 times
- Followed by:614 members
- GMAT Score:800
Without doing any math: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
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)!
Stuart Kovinsky | Kaplan GMAT Faculty | Toronto
Kaplan Exclusive: The Official Test Day Experience | Ready to Take a Free Practice Test? | Kaplan/Beat the GMAT Member Discount
BTG100 for $100 off a full course
-
- Master | Next Rank: 500 Posts
- Posts: 304
- Joined: Wed Jan 27, 2010 8:35 am
- Location: International Space Station
- Thanked: 11 times
- Followed by:3 members
-
- Senior | Next Rank: 100 Posts
- Posts: 46
- Joined: Sat Feb 27, 2010 2:58 am
- Location: GMAT
Love that approach !!!!!!!!!!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)!
GMAT, GMAT, GMAT, GMAT, GMAT, GMAT, GMAT, GMAT, GMAT, GMAT, GMAT, GMAT, GMAT, GMAT, GMAT, GMAT, GMAT, GMAT, GMAT, GMAT, GMAT, GMAT, GMAT, GMAT
What's life without GMAT !!!!!!!!
What's life without GMAT !!!!!!!!
-
- Junior | Next Rank: 30 Posts
- Posts: 29
- Joined: Thu Mar 11, 2010 3:23 am
- Thanked: 1 times
Me Too !!!GMATSUCKER wrote:Love that approach !!!!!!!!!!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)!