When two coins are tossed, then the sample space S contains 4 equally likely events. S = {HH, HT, TH, TT}
(i) E : ‘tail appears on one coin’and F : ‘one coin shows head’
⇒ E = {HT, TH} and F = {TH, HT}
⇒ E ∩ F = {HT, TH}
∴P(E)=

2

4

=

1

2

,P(F)=

2

4

=

1

2

and P(E∩F)=

2

4

=

1

2

∴P(E|F)=

P(E∩F)

P(F)

=

1 2

1 2

=1
(ii) E : ‘no tail appears’ and F : ‘no head appears’
⇒ E = {HH} and F = {TT} ⇒ E ∩ F =ϕ
Hence, P(E ∩ F) = 0 and P(F) = 1/4.
∴P(E|F)=

P(E∩F)

P(F)

=

0

1 4

=0