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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. \(\frac{1}{4}\)
A. \(\frac{1}{16}\)
B. \(\frac{3}{32}\)
C. \(\frac{1}{8}\)
D. \(\frac{3}{16}\)
E. \(\frac{1}{4}\)
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
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
11 Nov 2016, 14:46
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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
The last three throw should be HTH
There is only one way of getting this out of total 8 possibilities, so probability = 1/8
