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M21-11

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Bunuel
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M21-11 [#permalink] New post   16 Sep 2014, 01:10
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If a coin is tossed until the sequence \(HTH\) appears (\(H\) denotes the fall of heads, \(T\) denotes the fall of tails), what is the probability that the game will end with 4th toss?

A. \(\frac{1}{16}\)
B. \(\frac{3}{32}\)
C. \(\frac{1}{8}\)
D. \(\frac{3}{16}\)
E. \(\frac{1}{4}\)
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C
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Bunuel
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Re M21-11 [#permalink] New post   16 Sep 2014, 01:10
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Official Solution:

If a coin is tossed until the sequence \(HTH\) appears (\(H\) denotes the fall of heads, \(T\) denotes the fall of tails), what is the probability that the game will end with 4th toss?

A. \(\frac{1}{16}\)
B. \(\frac{3}{32}\)
C. \(\frac{1}{8}\)
D. \(\frac{3}{16}\)
E. \(\frac{1}{4}\)

We look for either \(HHTH\) or \(THTH\) sequence. The probability of either is \((\frac{1}{2})^4 = \frac{1}{16}\). The combined probability is \(\frac{2}{16} = \frac{1}{8}\).

Answer: C
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thapliya
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Re: M21-11 [#permalink] New post   11 Nov 2016, 14:46
Basically it does not matter what we get in the first throw.
The last three throw should be HTH
There is only one way of getting this out of total 8 possibilities, so probability = 1/8
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Re: M21-11 [#permalink] New post   14 Jun 2023, 00:40
How can we confirm if the game ends in the 4th toss if the number of heads & tails are equal? Would that not entail another toss, thereby meaning that the probability that the game ends on the 4th toss is 1/16 only? Kindly do clarify. Thank you!
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Re: M21-11 [#permalink] New post   14 Jun 2023, 00:47
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okrahulnambiar wrote:
How can we confirm if the game ends in the 4th toss if the number of heads & tails are equal? Would that not entail another toss, thereby meaning that the probability that the game ends on the 4th toss is 1/16 only? Kindly do clarify. Thank you!


I think you are missing the point. The game ends as soon as we get HTH sequence. The game to end at the fourth toss, we should have either H-HTH or T-HTH sequences.
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Re: M21-11 [#permalink]
14 Jun 2023, 00:47
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Bunuel
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