Mathematics

In a simultaneous throw of two coins, the probability of getting at least one head is:
$\Big(\dfrac{1}{2}\Big)$
$\Big(\dfrac{1}{3}\Big)$
$\Big(\dfrac{2}{3}\Big)$
$\Big(\dfrac{3}{4}\Big)$

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When two coins are thrown simultaneously, the possible outcomes are: {(H, H), (H, T), (T, H), (T, T)}

Total number of outcomes = 4

Let E be the event of getting at least one head,

E = {(H, H), (H, T), (T, H)}

The number of favorable outcomes to the event E = 3

∴ P(E) = $\dfrac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \dfrac{3}{4}$

Hence, option 4 is the correct option.

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