Harpreet tosses two different coins simultaneously (say, one is of Re 1 and other of Rs 2).

Total no. of possible outcomes = 4 which are{HT, HH, TT, TH}

E ⟶ event of getting at least one head

No. of favourable outcomes = 3 {HT, HH, TH}

Probability, P(E) = (No.of favorable outcomes)/(Total no.of possible outcomes)

P(E) = 3/4

We write H for ‘head’ and T for ‘tail’. When two coins are tossed simultaneously, the possible outcomes are (H, H), (H, T), (T, H), (T, T), which are all equally likely. Here (H, H) means head up on the first coin (say on Re 1) and head up on the second coin (Rs 2). Similarly (H, T) means head up on the first coin and tail up on the second coin and so on.

The outcomes favourable to the event E, ‘at least one head’ are (H, H), (H, T) and (T, H).

So, the number of outcomes favourable to E is 3.

Therefore,

P(E)=3/4

i.e., the probability that Harpreet gets at least one head is 3/4.