Question

Determine P(E|F):Two coins are tossed once.(i)E:tail appears on one coin,F:one coin shows head.(ii)E:no tail appears,F:no head appears.

Answer 1

A coin tossed three times, i.e.,S = (TTT, HTT, THT, TTH, HHT, HTH, THH, HHH)
⇒n(S)=8

(i)E: heads on third toss
E=(TTH, HTH, THH, HHH)
⇒N(E)=4

P(E)=n(E)/n(S)
=4/8
=1/2

F: heads on first two tosse
F=(HHT,HHH)
⇒n(F)=2

P(F)=n(F)/n(S)
=2/8
=1/4

∴E∩F=(HHH)
⇒n(E∩F)=1

∴P(E∩F)=n(E∩F)/n(S)
=1/8
And, P(E|F)=P(E∩F)/P(F)
=1/8/2/8
=1/2

(ii)E:at least two heads
E=(HHT,HTH,THH,HHH)
⇒n(E)=4

P(E)=n(E)/n(S)
=4/8
=1/2

F: at most two heads F=(TTT,HTT,THT,TTH,HHT,HTH,THH)
⇒n(F)=7

P(F)=n(F)/n(S)
=7/8

∴E∩F=(HHT,HTH,THH)
⇒n(E∩F)=3

∴P(E∩F)=n(E∩F)/n(S)
=3/8
And,
P(E|F)=P(E∩F)/P(F)
=3/8/7/8
=3/7

(iii)E: at most two tails E=(HTT,THT,TTH,HHT,HTH,THH,HHH)
⇒n(E)=7

P(E)=n(E)/n(S)
=7/8

F: at least one tail F=(TTT,HTT,THT,TTH,HHT,HTH,THH)
⇒n(F)=7

P(F)=n(F)/n(S)
=7/8

∴E∩F=(HTT, THT, TTH, HHT, HTH, THH)
⇒n(E∩F)
=6

∴P(E∩F)=n(E∩F)/n(S)
=6/8
And, P(E|F)=P(E∩F)/P(F)
=6/8/7/8
=6/7