A coin is tossed two times. The probability of getting atleast one tail is :

1. $\dfrac{3}{4}$

2. $\dfrac{4}{3}$

3. $\dfrac{1}{2}$

4. $\dfrac{5}{6}$

When a coin is tossed, each toss has 2 possible outcomes : head (H) or tail (T).

The number of possible outcomes when a coin is tossed n times = 2ⁿ

For this case, n = 2

So, 2² = 4

The number of possible outcomes when a coin is tossed 2 times is 4.

Number of favourable outcomes (Getting at least one tail) = 3 (HT, TH, TT)

P(Getting at least one tail) = $\dfrac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}$

= $\dfrac{3}{4}$

Hence, option 1 is the correct option.