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

When tossing 3 coins simultaneously, X represents the number of heads obtained.

The condition $0 < X < 3$ means X is more than 0 and less than 3, so X can be either 1 or 2.

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When 3 coins are tossed, each coin can be either Head (H) or Tail (T).

Total possible outcomes $= 2 \times 2 \times 2 = 8$

The complete list of outcomes with their corresponding number of heads:

HHH: 3 heads

HHT: 2 heads

HTH: 2 heads

HTT: 1 head

THH: 2 heads

THT: 1 head

TTH: 1 head

TTT: 0 heads

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For $0 < X < 3$, we need X = 1 or X = 2.

Outcomes where X = 1: HTT, THT, TTH (3 outcomes)

Outcomes where X = 2: HHT, HTH, THH (3 outcomes)

Total favorable outcomes $= 3 + 3 = 6$

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$P(0 < X < 3) = \dfrac{\text{Favorable outcomes}}{\text{Total outcomes}}$

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

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

Therefore, $P(0 < X < 3) = \dfrac{3}{4}$