Question

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

• A. \(\frac{21}{16}\)<
• B. \(\frac{15}{16}\)
• C. \(\frac{81}{64}\)
• D. \(\frac{37}{16}\)

Answer 1

The Correct Option is A

To solve this problem, we need to calculate the mean of the number of tosses until either a head or three tails appear using a biased coin. Here is the step-by-step solution:

Understanding the Problem

The coin is biased such that the probability of getting a head is 3 times the probability of getting a tail. Let the probability of getting a tail be \( P(T) = p \). Then, the probability of getting a head, \( P(H) \), is \( 3p \).

Since the total probability is 1, we have:

P(H) + P(T) = 1
3p + p = 1 \implies 4p = 1 \implies p = \frac{1}{4}
P(T) = \frac{1}{4}, \quad P(H) = 3p = \frac{3}{4}

Mean Calculation

The experiment continues until we either get a head or get three consecutive tails. We need to compute the expected number of tosses, \( E(X) \).

We can define different states as follows:

• State 0: Start state or no tail yet. Expected number of tosses: \( E_0 \).
• State 1: One tail has appeared. Expected number of tosses: \( E_1 \).
• State 2: Two tails have appeared. Expected number of tosses: \( E_2 \).

Equations for Each State

We derive equations based on how we can move from one state to another:

1. From State 0, we can:
   • Get a head (\( P(H) = \frac{3}{4} \)) and stop: Toss is successful.
   • Get a tail (\( P(T) = \frac{1}{4} \)) and move to State 1.
   E_0 = 1 + \frac{3}{4}(0) + \frac{1}{4}E_1 = 1 + \frac{1}{4}E_1
2. From State 1, we can:
   • Get a head (\( P(H) = \frac{3}{4} \)) and stop.
   • Get a tail (\( P(T) = \frac{1}{4} \)) and move to State 2.
   E_1 = 1 + \frac{3}{4}(0) + \frac{1}{4}E_2 = 1 + \frac{1}{4}E_2
3. From State 2, we can:
   • Get a head (\( P(H) = \frac{3}{4} \)) and stop.
   • Get a tail (\( P(T) = \frac{1}{4} \)) and the sequence ends with the 3rd tail.
   E_2 = 1 + \frac{3}{4}(0) + \frac{1}{4}(0) = 1

Solving the System of Equations

Substitute \( E_2 = 1 \) into \( E_1 = 1 + \frac{1}{4}E_2 \):

E_1 = 1 + \frac{1}{4} \times 1 = 1 + \frac{1}{4} = \frac{5}{4}

Now substitute \( E_1 \) into \( E_0 = 1 + \frac{1}{4}E_1 \):

E_0 = 1 + \frac{1}{4} \times \frac{5}{4} = 1 + \frac{5}{16} = \frac{16}{16} + \frac{5}{16} = \frac{21}{16}

Conclusion

Therefore, the mean of the number of tosses, \( E(X) \), is \(\frac{21}{16}\).