(i) The probability of failing to qualify on the first two jumps and succeeding on the third is
(1 - 0.8) (1 - 0.8) 0.8 = 0.032
In other words, if X is the random variable representing the first qualifying jump, then X is geometrically distributed and the probability of qualifying for the final round on exactly the n-th jump is
[tex]P(X=n) = (1-0.8)^{n-1}\times0.8[/tex]
(ii) In order to qualify for the final round, David has to succeed on at least one jump. Let Y be the random variable representing how many successful jumps are made. Then Y is binomially distributed with
[tex]P(Y = n) = \dbinom3n 0.8^n\times(1-0.8)^{3-n}[/tex]
The probability of succeeding at least once is
[tex]P(Y=1) + P(Y=2) + P(Y=3) \\\\ = \dbinom310.8(1-0.8)^2 + \dbinom320.8^2(1-0.8) + \dbinom33 0.8^3 = \boxed{0.992}[/tex]
