Question

A coin is tossed until one head appears or a tail appears \(4\) times in succession. The probability distribution of the number of tosses is

• A. A
• B. B
• C. C
• D. D

Hint:

In such stopping-time problems, list all possible terminal sequences clearly first. That makes the probability distribution easy to build.

Answer 1

The Correct Option is D

Concept:
Let \(X\) be the number of tosses. The experiment stops when either:
• the first head appears,
• or \(4\) tails appear consecutively e} So the possible values of \(X\) are: \[ 1,\ 2,\ 3,\ 4 \] ip

Step 1: Find \(P(X=1)\).
The process stops at the first toss if the first toss is head. \[ P(X=1)=\frac12 \] ip

Step 2: Find \(P(X=2)\).
This happens if the first toss is tail and the second toss is head: \[ TH \] So, \[ P(X=2)=\frac12\cdot\frac12=\frac14 \] ip

Step 3: Find \(P(X=3)\).
This happens if the first two tosses are tails and the third toss is head: \[ TTH \] Thus, \[ P(X=3)=\left(\frac12\right)^3=\frac18 \] ip

Step 4: Find \(P(X=4)\).
This happens in two ways:
• \(TTTH\)
• \(TTTT\) e} So, \[ P(X=4)=\left(\frac12\right)^4+\left(\frac12\right)^4 \] \[ P(X=4)=\frac{1}{16}+\frac{1}{16}=\frac18 \] ip

Step 5: Write the full probability distribution.
Therefore, \[ P(X=1)=\frac12,\quad P(X=2)=\frac14,\quad P(X=3)=\frac18,\quad P(X=4)=\frac18 \] ip Hence, choose the Option corresponding to:
\[ \boxed{ P(X=1)=\frac12,\ P(X=2)=\frac14,\ P(X=3)=\frac18,\ P(X=4)=\frac18 } \]