Bunuel wrote:
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. \(\frac{2}{16}\)
C. \(\frac{3}{16}\)
D. \(\frac{4}{16}\)
E. \(\frac{6}{16}\)
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 = \(\frac{1}{16} + \frac{1}{16} = \frac{2}{16}\).
Answer: B
Hi Bunuel,
Here's how i am approaching this question.
The question is asking - out of 4 throws, what's the probability of getting 3 consecutive heads or 3 consecutive tails.
Total # of arrangements = 2 * 2 * 2 * 2 = 16
Now, 3 consecutive heads can be like - THHH or HHHT
Similarly, 3 consecutive tails can be like - HTTT or TTTH.
So, total # of favourable outcomes = 2 + 2 = 4.
So, the probability = \(\frac{4}{16}\)
Can you please point to where i am wrong ?
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