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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