When two coins are tossed once, the sample space is given by
S={HH,HT,TH,TT}
and, therefore, n(S) = 4.
(i) Let E1
= event of getting 2 heads. Then,
E1={HH}
and, therefore, n(E1)=1
.
∴
P (getting 2 heads) =P(E1)=
n(E1)
n(S)
=
1
4
.
(ii) Let E2
= event of getting at least 1 head. Then,
E2={HT,TH,HH}
and, therefore, n(E2)=3
.
∴
P (getting at least 1 head) =P(E2)=
n(E2)
n(S)
=
3
4
.
(iii) Let E3
= event of getting no head. Then,
E3={TT}
and, therefore, n(E3)=1
.
∴
P (getting no head) =P(E3)=
n(E3)
n(S)
=
1
4
.
(iv) Let E4
= event of getting 1 head or 1 tail. Then,
E4={HT,TH}
and, therefore, n(E4)=2
.
∴
P (getting 1 head and 1 tail) =P(E4)=
n(E4)
n(S)
=
2
4
=
1
2
.
