Two coins are tossed once. Find the probability of
(i) getting 2 heads (ii) getting at least 1 head
(iii) getting no head (iv) getting 1 head and 1 tail

Explanation

When two coins are tossed once, the sample space is given by
`S = {HH, HT, TH, "TT"}` and, therefore, n(S) = 4.
(i) Let `E_(1)` = event of getting 2 heads. Then,
`E_(1) = {HH}` and, therefore, `n(E_(1)) = 1`.
`therefore ` P (getting 2 heads) `= P(E_(1)) = (n(E_(1)))/(n(S)) = 1/4.`
(ii) Let `E_(2)` = event of getting at least 1 head. Then,
`E_(2) = {HT, TH, HH}` and, therefore, `n(E_(2)) = 3`.
`therefore ` P (getting at least 1 head) `= P(E_(2)) = (n(E_(2)))/(n(S)) = 3/4.`
(iii) Let `E_(3)` = event of getting no head. Then,
`E_(3) = {"TT"}` and, therefore, `n(E_(3)) = 1`.
`therefore` P (getting no head) `= P(E_(3)) = (n(E_(3)))/(n(S)) = 1/4.`
(iv) Let `E_(4)` = event of getting 1 head or 1 tail. Then,
`E_(4) = {HT, TH}` and, therefore, `n(E_(4)) = 2`.
`therefore ` P (getting 1 head and 1 tail) `= P(E_(4)) = (n(E_(4)))/(n(S)) = 2/4 = 1/2.`