Let $X$ denotes the number of heads in a simultaneous toss of three coins, then $P (0 < x < 3)$ is

$\frac{3}{4}$

The correct answer is Option (2) → $\frac{3}{4}$

Possible outcomes when three coins are tossed:

HHH, HHT, HTH, THH, HTT, THT, TTH, TTT

Number of total outcomes = 8

Let $X$ = number of heads

$X$ can take values 0, 1, 2, 3

$P(0 < X < 3)$ means $X = 1$ or $X = 2$

Favorable outcomes for $X=1$: HTT, THT, TTH → 3 outcomes

Favorable outcomes for $X=2$: HHT, HTH, THH → 3 outcomes

Total favorable = 6

$P(0 < X < 3) = \frac{6}{8} = \frac{3}{4}$

$P(0 < X < 3) = \frac{3}{4}$