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. \(\begin{array}{ccccc} \mathrm{X} & 1 & 2 & 3 & 4 \\ \mathrm{P}(\mathrm{X}=x) & \frac{1}{8} & \frac{1}{8} & \frac{1}{2} & \frac{1}{4} \end{array}\)
• B. \(\begin{array}{ccccc} \mathrm{X} & 1 & 2 & 3 & 4 \\ \mathrm{P}(\mathrm{X}=x) & \frac{1}{4} & \frac{1}{2} & \frac{1}{8} & \frac{1}{8} \end{array}\)
• C. \(\begin{array}{ccccc} \mathrm{X} & 1 & 2 & 3 & 4 \\ \mathrm{P}(\mathrm{X}=x) & \frac{1}{8} & \frac{1}{4} & \frac{1}{8} & \frac{1}{2} \end{array}\)
• D. \(\begin{array}{ccccc} \mathrm{X} & 1 & 2 & 3 & 4 \\ \mathrm{P}(\mathrm{X}=x) & \frac{1}{2} & \frac{1}{4} & \frac{1}{8} & \frac{1}{8} \end{array}\)

Hint:

List the ways the tossing can stop after 1, 2, 3 or 4 tosses and multiply the probabilities of 1/2.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept
Tossing stops at the first head, or after four tails in a row. X is the number of tosses, so X can be 1, 2, 3 or 4.

Step 2: X = 1, 2, 3
\(P(X=1)=P(H)=\dfrac12\), \(P(X=2)=P(TH)=\dfrac14\), \(P(X=3)=P(TTH)=\dfrac18\).

Step 3: X = 4
The fourth toss ends the experiment either way: TTTH or TTTT. So \(P(X=4)=\dfrac1{16}+\dfrac1{16}=\dfrac18\).

Step 4: Check
\(\dfrac12+\dfrac14+\dfrac18+\dfrac18=1\), so the probabilities for X = 1, 2, 3, 4 are \(\dfrac12,\dfrac14,\dfrac18,\dfrac18\).

Final Answer:
The probabilities for X = 1, 2, 3, 4 are 1/2, 1/4, 1/8, 1/8. \[ \boxed{\text{(D) }X=1,2,3,4\text{ with }P(X)=\dfrac{1}{2},\dfrac{1}{4},\dfrac{1}{8},\dfrac{1}{8}} \]