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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
(A) 2 (B) 3 (C) 4 (D) 5 (E) 6
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23 Sep 2006, 12:56
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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...
I think its the 2nd, am I right?
Working on it...
23 Sep 2006, 13:45
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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?
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?
or 
or 
!
