A coin is tossed until one head appears or a tail appears 4 times in succession. The probability…
X 1 2 3 4 $\mathrm{P}(\mathrm{X}=x)$ $\frac{1}{8}$ $\frac{1}{8}$ $\frac{1}{2}$ $\frac{1}{4}$ X 1 2 3 4 $\mathrm{P}(\mathrm{X}=x)$ $\frac{1}{4}$ $\frac{1}{2}$ $\frac{1}{8}$ $\frac{1}{8}$ X 1 2 3 4 $\mathrm{P}(\mathrm{X}=x)$ $\frac{1}{8}$ $\frac{1}{4}$ $\frac{1}{8}$ $\frac{1}{2}$ X 1 2 3 4 $\mathrm{P}(\mathrm{X}=x)$ $\frac{1}{2}$ $\frac{1}{4}$ $\frac{1}{8}$ $\frac{1}{8}$
Solution
The process terminates when a head appears or four consecutive tails occur. The number of tosses, $X$, has possible values $1$, $2$, $3$, and $4$.
An outcome of a head on the first toss gives $X = 1$, with probability $P(X=1) = \frac{1}{2}$.
For $X = 2$, the sequence is $TH$, so $P(X=2) = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$.
For $X = 3$, the sequence is $TTH$, giving $P(X=3) = (\frac{1}{2})^3 = \frac{1}{8}$.
$X = 4$ occurs either with $TTTH$ or $TTTT$, each with probability $(\frac{1}{2})^4 = \frac{1}{16}$, so $P(X=4) = \frac{1}{16} + \frac{1}{16} = \frac{1}{8}$.
Summing the probabilities confirms $\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{8} = 1$.
The probability distribution matches Option D.
Asked in: MHT CET 2025 (22 April Shift 1)