Find the probability distribution of the number of tails in the simultaneous tosses of three coins.

The correct answer is Option (1) → 

A simultaneous toss of three coins is the same as tossing of one coin thrice.

Probability of getting a tail = $p = \frac{1}{2}$, so $q = 1-\frac{1}{2}=\frac{1}{2}$

A coin is tossed thrice, events are independent, therefore, it is a problem of binomial distribution with $p=\frac{1}{2},q = \frac{1}{2}$ and $n = 3$.

If X denotes the number of tails, then X can take values 0, 1, 2, 3.

$P(0)={^3C}_0q^3=1.(\frac{1}{2})^3=\frac{1}{8},$

$P(1)={^3C}_1pq^2=3.\frac{1}{2}.(\frac{1}{2})^3=\frac{3}{8},$

$P(2)={^3C}_2p^2q=3.(\frac{1}{2})^2.\frac{1}{2}=\frac{3}{8}$ and

$P(3)={^3C}_3p^3=1.(\frac{1}{2})^3=\frac{1}{8}$

∴ The probability distribution of number of tails in the simultaneous tosses of three coins is $\begin{pmatrix}0&1&2&3\\\frac{1}{8}&\frac{3}{8}&\frac{3}{8}&\frac{1}{8}\end{pmatrix}$.