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A bag contains 10 coins, 9 of which are fair coins. The

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kevincan
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A bag contains 10 coins, 9 of which are fair coins. The [#permalink] New post   Updated on: 25 Sep 2006, 07:51
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A bag contains 10 coins, 9 of which are fair coins. The remaining coin has two heads. Marcia removes a coin and tosses it repeatedly, and to her surprise, it turns up heads after every toss. How many times will Marcia have to toss this coin until the probability that the coin has two heads is greater than 0.5?

(A) 2 (B) 3 (C) 4 (D) 5 (E) 6



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Originally posted by kevincan on 22 Sep 2006, 14:40.
Last edited by kevincan on 25 Sep 2006, 07:51, edited 2 times in total.
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Re: A bag contains 10 coins, 9 of which are fair coins. The [#permalink] New post   23 Sep 2006, 12:56
Does this mean that the coin is removed once and tossed repeatedly? Or does this mean that the coin is removed, tossed, and put back...

I think its the 2nd, am I right?

Working on it... :roll:
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Re: A bag contains 10 coins, 9 of which are fair coins. The [#permalink] New post   23 Sep 2006, 13:45
Hmm...

Take P(t) = Probability of trick coin
P(f) = Probability of fair coin

We know that we always get heads.
Hence, 1 = P(t)*(1^n) + P(f)*[(0.5)^n]
=> 1 = P(t) + P(f)*(0.5)^n

Max. possible value of P(f) is 1, since P(f) + P(t) = 1. So with one throw we cannot guarantee that P(t) > 0.5, but with 2 throws we can.

Answer: A

Reasoning seems to be icky... Not confident at all with my work! what do you say?
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Re: A bag contains 10 coins, 9 of which are fair coins. The [#permalink] New post   26 Sep 2006, 12:54
paddyboy wrote:
Hmm...

Take P(t) = Probability of trick coin
P(f) = Probability of fair coin

We know that we always get heads.
Hence, 1 = P(t)*(1^n) + P(f)*[(0.5)^n]
=> 1 = P(t) + P(f)*(0.5)^n

Max. possible value of P(f) is 1, since P(f) + P(t) = 1. So with one throw we cannot guarantee that P(t) > 0.5, but with 2 throws we can.

Answer: A

Reasoning seems to be icky... Not confident at all with my work! what do you say?


The problem with this is that it does not take into consideration the probability of picking a coin initially... :roll: and can't seem to figure it out. Any hints?
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kevincan
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Re: A bag contains 10 coins, 9 of which are fair coins. The [#permalink] New post   26 Sep 2006, 13:01
kevincan wrote:
A bag contains 10 coins, 9 of which are fair coins. The remaining coin has two heads. Marcia removes a coin and tosses it repeatedly, and to her surprise, it turns up heads after every toss. How many times will Marcia have to toss this coin until the probability that the coin has two heads is greater than 0.5?

(A) 2 (B) 3 (C) 4 (D) 5 (E) 6


Let H be the event that the coin chosen has two heads and Q be the event that n heads in a row are obtained.

What is Pr(H/Q) ? (This concept is in the OG Quantitative Review pg 29, so don't give me any :( or :? or :roll: !



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Re: A bag contains 10 coins, 9 of which are fair coins. The [#permalink]
26 Sep 2006, 13:01
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