Question

A coin is biased so that a head is twice as likely to occur as a tail. If the coin is tossed \(3\) times, what is the probability of getting \(2\) tails and \(1\) head.

• A. \(2/9\)
• B. \(1/9\)
• C. \(2/27\)
• D. \(4/27\)

Hint:

If head is twice as likely as tail, write: \[ P(T)=x,\quad P(H)=2x \] then use \(x+2x=1\).

Answer 1

The Correct Option is A

Concept:
This is a binomial probability question. For exactly \(r\) successes in \(n\) trials: \[ P(X=r)={}^{n}C_r p^r q^{n-r} \] Here, we need exactly \(2\) tails and \(1\) head in \(3\) tosses.

Step 1: Find probabilities of head and tail.
It is given that head is twice as likely as tail. Let: \[ P(T)=x \] Then: \[ P(H)=2x \] Since total probability is \(1\): \[ x+2x=1 \] \[ 3x=1 \] \[ x=\frac13 \] So: \[ P(T)=\frac13 \] and: \[ P(H)=\frac23 \]

Step 2: Write the required event.
We need: \[ 2\text{ tails and }1\text{ head} \] The number of ways to arrange \(2\) tails among \(3\) tosses is: \[ {}^3C_2=3 \]

Step 3: Calculate probability.
\[ P={}^{3}C_2\left(\frac13\right)^2\left(\frac23\right) \] \[ P=3\cdot\frac19\cdot\frac23 \] \[ P=\frac{3\times2}{9\times3} \] \[ P=\frac{2}{9} \]

Step 4: Check the options.
Option (A) \(2/9\) is correct.
Option (B) \(1/9\) is incorrect.
Option (C) \(2/27\) is missing the arrangement factor.
Option (D) \(4/27\) is not correct. Hence, the correct answer is: \[ \boxed{(A)\ \frac{2}{9}} \]