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11 Dec 2012, 12:59
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B
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Difficulty:
45%
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Question Stats:
65% (01:20) correct
35%
(01:27)
wrong
based on 1331
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If John throws a coin until a series of three consecutive heads or three consecutive tails appears, what is the probability that the game will end on the fourth throw?
A. \(\frac{1}{16}\)
B. \(\frac{1}{8}\)
C. \(\frac{3}{16}\)
D. \(\frac{1}{4}\)
E. \(\frac{3}{8}\)
A. \(\frac{1}{16}\)
B. \(\frac{1}{8}\)
C. \(\frac{3}{16}\)
D. \(\frac{1}{4}\)
E. \(\frac{3}{8}\)
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19 Apr 2024, 05:28
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Official Solution:
If John throws a coin until a series of three consecutive heads or three consecutive tails appears, what is the probability that the game will end on the fourth throw?
A. \(\frac{1}{16}\)
B. \(\frac{1}{8}\)
C. \(\frac{3}{16}\)
D. \(\frac{1}{4}\)
E. \(\frac{3}{8}\)
Calculate the probability that either of the two sequences, \(THHH\) or \(HTTT\), will occur. The probability of \(THHH\) occurring = \((\frac{1}{2})^4 = \frac{1}{16}\). The probability of \(HTTT\) occurring = \((\frac{1}{2})^4 = \frac{1}{16}\). The probability of either \(THHH\) or \(HTTT\) occurring = \(\frac{1}{16} + \frac{1}{16} = \frac{2}{16} = \frac{1}{8}\).
Answer: B
If John throws a coin until a series of three consecutive heads or three consecutive tails appears, what is the probability that the game will end on the fourth throw?
A. \(\frac{1}{16}\)
B. \(\frac{1}{8}\)
C. \(\frac{3}{16}\)
D. \(\frac{1}{4}\)
E. \(\frac{3}{8}\)
Calculate the probability that either of the two sequences, \(THHH\) or \(HTTT\), will occur. The probability of \(THHH\) occurring = \((\frac{1}{2})^4 = \frac{1}{16}\). The probability of \(HTTT\) occurring = \((\frac{1}{2})^4 = \frac{1}{16}\). The probability of either \(THHH\) or \(HTTT\) occurring = \(\frac{1}{16} + \frac{1}{16} = \frac{2}{16} = \frac{1}{8}\).
Answer: B
General Discussion
11 Dec 2012, 13:09
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david1983
Great question! I'm happy to help.
Let's consider the routes that would lead to ending the game on the fourth toss. In order to be a win at that point, tosses #2 & #3 & #4 would all have to be the same, either H or T, and in order for the game NOT to end on the third toss, the first toss would have to be different from the other four. Thus, the only two possible sequences are
H-T-T-T ----> P = (1/2)^4 = 1/16
T-H-H-H ----> P = (1/2)^4 = 1/16
Either one of these would satisfy the condition, so we could have one or the other. OR means add in probability.
1/16 + 1/16 = 2/16 = 1/8
Answer (B). Incidentally, the GMAT would NEVER give the unsimplified fraction 2/16 --- it would always give the simplified form 1/8.
Does all this make sense?
Mike
