# 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.

## 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.

## To solve this problem, we can use the concept of probabilities and combinations.

(i) In Game 1, Darcy wins if he hits the target at least once. To calculate the probability of winning, we can find the probability of Darcy not hitting the target in both throws and subtract it from 1.

The probability of Darcy not hitting the target in one throw is (1 - p). Since Darcy throws two darts, the probability of not hitting the target in both throws is (1 - p)(1 - p) = (1 - p)^2.

Therefore, the probability of Darcy hitting the target in at least one throw (winning Game 1) is 1 - (1 - p)^2.

(ii) In Game 2, Darcy wins if he hits the target at least twice. To calculate the probability of winning, we need to consider the different combinations of hits and misses in the three throws.

The probability of Darcy hitting the target in one throw is p, and the probability of not hitting the target in one throw is (1 - p).

To calculate the probability of Darcy hitting the target exactly twice, we need to consider the three possible combinations: HHM, HMH, and MHH (H represents a hit, and M represents a miss). The probability of each of these combinations is p*p*(1-p) = p^2(1-p).

Since there are three possible combinations, the probability of Darcy hitting the target exactly twice is 3p^2(1-p).

To calculate the probability of Darcy hitting the target at least twice (winning Game 2), we need to also consider the possibility of hitting the target on all three throws. The probability of hitting the target in all three throws is p^3.

Therefore, the probability of Darcy hitting the target at least twice (winning Game 2) is 3p^2(1-p) + p^3.

(iii) To compare the probabilities of winning Game 1 and Game 2, we can compare their expressions:

Probability of winning Game 1: 1 - (1 - p)^2 = 2p - p^2.

Probability of winning Game 2: 3p^2(1-p) + p^3.

To show that Darcy is more likely to win Game 1 than Game 2, we need to show that the probability of winning Game 1 is greater than the probability of winning Game 2 for all values of p.

By comparing the two expressions, it is apparent that 2p - p^2 is a quadratic function with a negative quadratic term, while 3p^2(1-p) + p^3 is a cubic function with a positive cubic term.

Since a quadratic function with a negative quadratic term is always greater than a cubic function with a positive cubic term, we can conclude that Darcy is more likely to win Game 1 than Game 2.

(iv) To find the value of p for which Darcy is twice as likely to win Game 1 as he is to win Game 2, we can set up the following equation:

2p - p^2 = 2(3p^2(1-p) + p^3).

Simplifying the equation, we have:

2p - p^2 = 6p^2 - 2p^3 + 2p^3.

Rearranging the terms and combining like terms, we get:

p^3 - 4p^2 + 2p = 0.

Factoring out p, we have:

p(p^2 - 4p + 2) = 0.

Setting each factor equal to zero, we can solve for p:

p = 0 (not possible since p > 0),

or p^2 - 4p + 2 = 0.

Using the quadratic formula, we can solve for p:

p = (4 ± sqrt(4^2 - 4(1)(2)) / (2(1)).

p = (4 ± sqrt(16 - 8)) / 2.

p = (4 ± 2sqrt(2)) / 2.

p = 2 ± sqrt(2).

Therefore, Darcy is twice as likely to win Game 1 as he is to win Game 2 when p = 2 + sqrt(2) or p = 2 - sqrt(2).