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19 Sep 2019, 10:38
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Difficulty:
25%
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Question Stats:
80% (01:00) correct
20%
(01:09)
wrong
based on 40
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Two children are playing a game that involves taking alternative turns in tossing a coin. The game ends when one of the children first obtains a "head" on their toss of the coin. If the game started with the first child tossing the coin, then what is the probability that this child will win the game?
(A) \(\frac{1}{6}\)
(B) \(\frac{1}{4}\)
(C) \(\frac{1}{3}\)
(D) \(\frac{1}{2}\)
(E) \(\frac{2}{3}\)
(A) \(\frac{1}{6}\)
(B) \(\frac{1}{4}\)
(C) \(\frac{1}{3}\)
(D) \(\frac{1}{2}\)
(E) \(\frac{2}{3}\)
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19 Sep 2019, 10:51
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I believe the answer is E. But not sure!
19 Sep 2019, 23:57
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Let, the 2 boys tossing coins be A and B, A starts first. Let p be the probability that A wins. This can happen in two ways: (i) A wins immediately (probability 1/2) or (ii) A tosses a tail, but ultimately wins.
If A tossed a tail (probability 1/2 , then in effect B is now "first" so the probability she does not win is (1-p) . We conclude that
p = 1/2 + 1/2 (1-p)
Solve for p. We get p = 2/3
If A tossed a tail (probability 1/2 , then in effect B is now "first" so the probability she does not win is (1-p) . We conclude that
p = 1/2 + 1/2 (1-p)
Solve for p. We get p = 2/3
