Question

A coin is tossed three times. If X denotes the absolute difference between the number of heads and the number of tails then $P\left(X=1\right)$ is

• A. $1/2$
• B. $1/6$
• C. $2/3$
• D. $3/4$

Answer 1

The correct option is D

$3/4$

We know that, when a coin is tossed $3$ times.

The total number of cases = $8$

Then, Sample space = {HHH, HTH, HHT, THH, TTT, TTH, THT, THT}

Now, X denotes the absolute difference between the number of heads and the number of tails i.e. $P\left(X=1\right)$. It means let's take an example of HTH, Here, Heads are $2$ and tails are $1$, So the difference between them is $2-1=1$. So it follows the condition. Now, we will find the same favourable conditions.

For $X=1$, favourable cases are HTH, HHT, THH, TTH, THT, THT.

Number of favourable cases = $6$

Now, $Requiredprobability=\frac{favourablecases}{Samplespace}$

$Requiredprobability=\frac{6}{8}=\frac{3}{4}$

Hence option D is the answer.