When a coin is tossed 4 times. A total of 2⁴ = 16 outcomes are possible.

Let S be the set consisting of all such outcomes.

∴ n(S) = 16

Let A be the event of getting 2 tails.

∴ A = { TTHH,THTH,THHT,HTTH,HTHT,HHTT}

∴ n(A) = 6

∴ P(A) = 6/16 = 3/8

Let B be the event of getting 3 tails.

∴ B = { TTTH ,TTHT, THTT,HTTT }

⇒ n(B) = 4

∴ P(B) = 4/16 = 1/4

We need to find the probability of getting 2 tails or 3 tails i.e.

P(A ∪ B) = ?

As we can’t get 2 and 3 tails at the same time. So A and B are mutually exclusive events.

∴ P(A ∪ B) = P(A) + P(B) = 3/8 + 1/4 = 5/8