One coin is tossed 3 times so total number of favourable outcomes `= 2^(3) = 8`, which are (HHH),(HHT), (THH) and (replacing ` H to T and T to H`) (TTT),(TTH),(THT),(HTT).
(i) Losing the game means getting no head.
Number of favourable outcomes of getting no head = F(E) = 1
So, `= P(E) = (F(E))/(T(E)) = (1)/(8)`
So, probability (losing the entry fee) i.e., `P(E) = (F(E))/(T(E)) = (1)/(8)` ltbr. (ii) Gets double entry fee back by getting 3 heads.
Number of favourable outcomes of getting 3 heads i.e, F(E) = 1
`therefore` Probability (getting double entry fee ) P(E) `= (F(E))/(T(E)) = (1)/(8)`
(iii) Just gets her enrty fees back by getting either one or two head,
Number of favouarble outcomes of getting either one or two heads, i.e, F(E) =6
`therefore` Probability (just getting entry fee) i.e.,`P(E) = (F(E))/(T(E)) = (6)/(8) =(3)/(4)`