Math

A game is played by throwing darts at a target. A player can choose to throw two or three darts.

Darcy plays two games. In Game 1, he chooses to throw two darts, and wins if he hits the target at least once. In Game 2, he chooses to throw three darts, and wins if he hits the target at least twice.

The probability that Darcy hits the target on any throw is p, where 0 < p < 1.

(i) Show that the probability that Darcy wins Game 1 is 2p - p[squared].

(ii) Show that the probaility that Darcy wins Game 2 is 3p[squared] - 2p[cubed].

(iii) Prove that Darcy is more likely to win Game 1 than Game 2.

(iv) Find the value of p for which Darcy is twice as likely to wine Game 1 as he is to win Game 2.

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  1. pr(win)=1-pr(lose)=1-(1-p)(1-p)

    ii pr (win)=p*P(1-P)

    iii is 2P-P^2>p^2-p^3 ?
    divide by p
    2-p>? P-P^2

    2-2p+P^2>?0
    (2-P)(1-P)>?0
    because p is between zero and 1, then each of the two terms on the left is positive, so the left side is indeed greater than zero.

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    bobpursley

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