Question

A biased coin with probability \(p\) (where \(0 < p < 1\)) of getting head is tossed until a head appears for the first time. If the probability that the number of tosses required is even is \(\frac{2}{5}\), then \(p =\):

• A. \(\frac{1}{4}\)
• B. \(\frac{1}{3}\)
• C. \(\frac{2}{3}\)
• D. \(\frac{3}{4}\)

Hint:

For problems involving geometric series, use the formula for the sum of an infinite series to calculate probabilities.

Answer 1

The Correct Option is B

The probability of the first head appearing on the \( k \)-th toss is:

\[ P(\text{first head on toss } k) = (1-p)^{k-1} \cdot p \]

This is because the first \( k-1 \) tosses must be tails (probability \( 1-p \)), and the \( k \)-th toss must be heads (probability \( p \)).

Step 2: We need to find the probability that an even number of tosses is required. This corresponds to the sum of probabilities for even values of \( k \) (i.e., \( k = 2, 4, 6, \ldots \)).

Step 3: The total probability of the first head occurring on an even toss is:

\[ P(\text{even toss}) = (1-p)p + (1-p)^3p + (1-p)^5p + \cdots \]

This is an infinite geometric series. The first term is \( (1-p)p \), and the common ratio is \( (1-p)^2 \).

Step 4: The sum of this infinite geometric series is:

\[ P(\text{even toss}) = \frac{(1-p)p}{1 - (1-p)^2} \]

Simplifying the denominator:

\[ 1 - (1-p)^2 = 1 - \left(1 - 2p + p^2\right) = 2p - p^2 \]

Therefore, the probability of an even toss is:

\[ P(\text{even toss}) = \frac{(1-p)p}{2p - p^2} \]

Step 5: We are given that \( P(\text{even toss}) = \frac{2}{5} \). Thus:

\[ \frac{(1-p)p}{2p - p^2} = \frac{2}{5} \]

Step 6: Cross-multiplying and simplifying:

\[ 5(1-p)p = 2(2p - p^2) \] \[ 5p - 5p^2 = 4p - 2p^2 \] \[ 5p - 4p = 5p^2 - 2p^2 \] \[ p = 3p^2 \] \[ 3p^2 - p = 0 \] \[ p(3p-1) = 0 \]

Step 7: Solving for \( p \), we find \( p = 0 \) or \( p = \frac{1}{3} \). Since \( p = 0 \) is not a valid probability, we conclude:

\[ p = \frac{1}{3} \]