A coin is biased so that the head is 3 times as likely to occur as tail. This coin is tossed until a head or three tails occur. If X denotes the number of tosses of the coin, then the mean of X is
- \(\frac{21}{16}\)<
- \(\frac{15}{16}\)
- \(\frac{81}{64}\)
- \(\frac{37}{16}\)
The Correct Option is A
Solution and Explanation
Since the coin is biased so that the head (H) is 3 times as likely to occur as tail (T), we have \[ P(\text{H}) = \frac{3}{4} \quad \text{and} \quad P(\text{T}) = \frac{1}{4}. \] Let \(X\) be the number of tosses until we get the first head or three tails in total. Clearly, \(X\) can only be \(1\), \(2\), or \(3\).
\(X = 1\) if we get a head on the first toss.
\[ P(X=1) = P(\text{H on first toss}) = \frac{3}{4}. \]
\(X = 2\) if the first toss is a tail and the second toss is a head.
\[ P(X=2) = P(\text{T}) \cdot P(\text{H}) = \frac{1}{4} \times \frac{3}{4} = \frac{3}{16}. \]
\(X = 3\) if the first two tosses are tails and then either the third toss is a head or it is a tail (making three tails in total).
\[ P(X=3) = P(\text{T}\,\text{T}\,\text{H}) + P(\text{T}\,\text{T}\,\text{T}) = \left(\frac{1}{4}\right)^2 \cdot \frac{3}{4} + \left(\frac{1}{4}\right)^3 = \frac{3}{64} + \frac{1}{64} = \frac{4}{64} = \frac{1}{16}. \]
These events cover all possibilities and sum to 1: \[ \frac{3}{4} + \frac{3}{16} + \frac{1}{16} = 1. \] The mean or expected value of \(X\) is then \[ \mathbb{E}[X] = 1 \cdot \frac{3}{4} + 2 \cdot \frac{3}{16} + 3 \cdot \frac{1}{16} = \frac{3}{4} + \frac{6}{16} + \frac{3}{16} = \frac{3}{4} + \frac{9}{16} = \frac{12}{16} + \frac{9}{16} = \frac{21}{16}. \] \[ \boxed{ \mathbb{E}[X] = \frac{21}{16}. } \]
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