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.

To solve this problem, we can use the concept of complementary probabilities. We will find the probability of Darcy not winning each game and subtract it from 1 to find the probability of Darcy winning.

(i) Probability that Darcy does not win Game 1:
In a single throw, the probability of not hitting the target 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.
Thus, the probability that Darcy does not win Game 1 is (1 - (1 - p)^2).

The probability that Darcy wins Game 1 is:
1 - (1 - (1 - p)^2)
= 2p - p^2

(ii) Probability that Darcy does not win Game 2:
In a single throw, the probability of not hitting the target is (1 - p). Since Darcy throws three darts, the probability of hitting the target less than twice is the sum of the following three cases:
1. Hitting the target once and missing twice: p * (1 - p) * (1 - p) = p(1 - p)^2
2. Missing the target once and hitting twice: (1 - p) * p * p = p^2(1 - p)
3. Missing the target all three times: (1 - p)^3

Thus, the probability that Darcy does not win Game 2 is: p(1 - p)^2 + p^2(1 - p) + (1 - p)^3.

The probability that Darcy wins Game 2 is:
1 - [p(1 - p)^2 + p^2(1 - p) + (1 - p)^3]
= 3p^2 - 2p^3

(iii) To prove 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. From (i) and (ii), we have:
2p - p^2 > 3p^2 - 2p^3

Simplifying this inequality, we get:
2 - p > 3p - 2p^2

Rearranging, we have:
2 - 3p > -2p^2 + p
2 + p - 3p > -2p^2
2 + p(1 - 3) > -2p^2
2 - 2p > -2p^2

Dividing both sides by -2, we get:
p^2 - p + 1 > 0

The equation p^2 - p + 1 > 0 holds true for all values of p, since the discriminant (b^2 - 4ac) is negative. Therefore, the probability of Darcy winning Game 1 is always greater than the probability of winning 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 set the probabilities from (i) and (ii) equal and solve for p:
2p - p^2 = 2(3p^2 - 2p^3)

Expanding the equation and rearranging terms, we get:
2p - p^2 = 6p^2 - 4p^3
4p^3 - 6p^2 + p^2 - 2p = 0
4p^3 - 5p^2 - 2p = 0

Factoring out p, we have:
p(4p^2 - 5p - 2) = 0

Setting each factor equal to zero:
p = 0 (not a valid solution since 0 < p < 1)

Solving the quadratic factor:
4p^2 - 5p - 2 = 0

Using factoring or the quadratic formula, we find the solutions for p:
p = 2/4 or p = -1/2 (not a valid solution since 0 < p < 1)
p = -1/4 or p = 2/1 (not valid solutions)

Therefore, there is no value of p for which Darcy is twice as likely to win Game 1 as he is to win Game 2.

To find the probabilities that Darcy wins Game 1 and Game 2, we can consider the complementary probabilities - the probabilities that Darcy does not hit the target in each game.

(i) In Game 1, Darcy wins if he hits the target at least once. The probability that he does not hit the target on any of the two throws is equal to (1-p) * (1-p) = (1-p)^2. Therefore, the probability that Darcy wins Game 1 is 1 - (1-p)^2.

(ii) In Game 2, Darcy wins if he hits the target at least twice. The probability that he does not hit the target on any of the three throws is equal to (1-p) * (1-p) * (1-p) = (1-p)^3. Therefore, the probability that Darcy wins Game 2 is 1 - (1-p)^3.

(iii) To compare the two probabilities, we need to compare 1 - (1-p)^2 and 1 - (1-p)^3. Let's simplify both expressions:
1 - (1-p)^2 = 1 - (1 - 2p + p^2) = 2p - p^2.

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

To prove that Darcy is more likely to win Game 1 than Game 2, we need to show that 2p - p^2 > 1 - (1 - 3p + 3p^2 - p^3).

Expanding the above expression, we get:
2p - p^2 > 1 - 1 + 3p - 3p^2 + p^3
2p - p^2 > 3p - 3p^2 + p^3.

Rearranging the inequality, we have:
p^3 - 2p^2 + p > 0.

We can factor out p from the expression:
p(p^2 - 2p + 1) > 0.

Simplifying further gives us:
p(p-1)^2 > 0.

Since p is a probability between 0 and 1, p > 0 and p-1 < 0. Therefore, (p-1)^2 > 0.

Since both terms are positive, the product p(p-1)^2 will also be positive. This implies 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 need to equate the probabilities obtained in (i) and (ii).

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

Simplifying the equation gives us:
2p - p^2 = 6p^2 - 4p^3.

Rearranging and simplifying further:
4p^3 - 6p^2 + 2p = 0.

Factoring out 2p, we get:
2p(2p^2 - 3p + 1) = 0.

Since p > 0, we can divide the equation by 2p, giving us:
2p^2 - 3p + 1 = 0.

This quadratic equation can be factored as:
(2p - 1)(p - 1) = 0.

Therefore, p = 1/2 or p = 1.

Checking both values in the original equation, we find that p = 1/2 is the solution since it satisfies the condition that Darcy is twice as likely to win Game 1 as he is to win Game 2.