A coin is tossed three times. Find `P(A//B)` in each of the following: `A=` Heads on third toss, B=Heads on first tow tosses `A=` At least two heads, B=At most two heads `A=` At most two tails, B=At least one tail.

Answer

N/a

Explanation

A coin is tossed three times, then the sample space of the experiment.
`S={HHH,HHT,HTH,H"TT",THH,THT,"TT"H,"TTT"}`
(i) Here `E=` event of getting head in 3rd throw and
`F=` event of gettng head in Ist and Iind throw.
`E={"HHH,THHltHTH,TT"H}`
and `F={HHH,HHT}`
`=EnnF={HHH}`
Therfore `P(E)\4/8=1/2`
`P(F)=2/8=1/4` and `P(EnnF)=1/8`
`:. P(E/F)=(P(nnF))/(P(F))`
`implies P(E/F)=(1/8)/(1/4)=1/8xx4/1=1/2`
(ii) Here `E=` event of getting at least two heads and
`F=` event of getting at most one head.
`E={HHH,HHT,HTH,THH}`
`F={"TTT",THT,"TT"H,H"TT"}`
and `(EnnF)=phi`
Therefore `P(E)=4/8=1/2`
`P(F)=4/8=1/2` and `P(EnnF)=0`
`:.P(E/F)=(P(EnnF))/(P(F))=0`
(iii) Here `E=` event of getting at most two tails and `F=` event of getting at least two tails.
`:. E={HHH,HHT,HTH,THH,H"TT",THT,"TTH}`
and `F={"TTT",THT,"TT"H,H"TT"}`
`:.(EnnF)={H"TT",THT,"TT"H}`
Now `P(E)=7/8`
`P(F)=4/8` and `P(EnnF)=3/8`
`:.P(E/F)=(P(EnnF))/(P(F))=(3//8)/(4//8)=3/4`