Two children are playing a game that involves taking alternative turns

Question

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}

StanLeRoi612 · Answer

1/2

Answer : D

IanStewart · Answer

The "OA" is listed as "D" right now, which obviously isn't right, because the first child has a 1/2 probability of winning on the first coinflip alone, but can also win later. So the answer must be greater than 1/2, and only E could conceivably be right.

To solve this kind of question properly though, you need to calculate an infinite sum, and you never need to do that on the GMAT.

Carcass · Answer

I think  IanStewart  you di not notice that the question is archived.

However,m the reply from you is still on point

Regards

Two children are playing a game that involves taking alternative turns

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