A coin is tossed and a die is thrown. The probability that the outcome will be a tail on the coin or a number greater than 3 on the die is

$\frac{3}{4}$

The correct answer is Option (1) → $\frac{3}{4}$

Total outcomes when a coin is tossed = 2 (Head, Tail)

Total outcomes when a die is thrown = 6 (1, 2, 3, 4, 5, 6)

Total outcomes of the combined experiment = $2 \times 6 = 12$

Let A be the event that a tail appears on the coin.

Then, outcomes favorable to A = {Tail with 1, 2, 3, 4, 5, 6} = 6 outcomes

Let B be the event that a number > 3 appears on the die.

Then, outcomes favorable to B = {Head with 4, 5, 6 and Tail with 4, 5, 6} = 6 outcomes

Now, A ∩ B = outcomes where both tail appears and die shows number > 3 = {Tail with 4, 5, 6} = 3 outcomes

So, by addition rule:

$P(A \cup B) = P(A) + P(B) - P(A \cap B)$

$= \frac{6}{12} + \frac{6}{12} - \frac{3}{12} = \frac{9}{12} = \frac{3}{4}$