When a coin is tossed three times, the sample space is given by
`S={HHH, HHT, HTH, THH, T TH, THT, HT T, T T T}`.
Now, `E=` event that the first throw results in a head.
`:. E={HHH, HHT, HTH, HT T}`.
And, `F=` event that the last throw results in a tail.
`:. F={HHT, THT, HT T, T T T}`.
So, `(E nn F)={HHT, HT T}`.
Clearly, `n(E)=4, n(F)=4, n(E nn F)=2` and `n(S)=8`.
`:. P(E)=(n(E))/(n(S))=4/8=1/2, P(F)=(n(F))/(n(S))=4/8=1/2`
and `P(E nn F)=(n(E nn F))/(n(s))=2/8=1/4`
thus, `P(E nn F)=P(E)xxP(F)`
Hence, E and F are independent events.