A coin is tossed three times, where
(i) A : head on third toss,B: heads on first two tosses
find P(A/B).

Explanation

If a coin is tossed three times, then the sample S is
`S={HHH,HHT, HTH, HT T ,THH,THT, T TH, T T T }`
`thereforen(S)=8.`
(i) `A={HHH, HTH, THH, T TH]`
` B= {HHH,HHT}`
`thereforeAnnB={HHH}`
`P(B)=2/8=1/4and P(AnnM)=1/8`
`P(A//B)=(P(AnnB))/(P(B))=(1/8)/(1/4)=4/8=1/2`
(ii) `A={HHH, HHT, HTH, THH}`
`B={HHT,HTH, HT T,THH,THT, T TH, T T T}`
`thereforeAnnB ={HHT, HTH, THH}`
Clearly, `P(AnnB)=3/8andP(B)=7/8`
`thereforeP(A//B)=(P(AnnB))/(P(B))=(3/8)/(7/8)=3/7`
(iii) `A={HHH, HHT, HT T , HTH, THH, THT, T T H}`
` B={HHT, HT T,HTH, THH, THT, T TH, T T T}`
`thereforeAnnB={HHT, HT T,HHT,THH,THT, T TH}`
`P(B)=7/8and P(AnnB)=6/8`
Therefore, `P(A//B)=(P(AnnB))/(P(B))=(6/8)/(7/8)=6/7`