Harry shoots arrows at a target board - NSC Mathematics - Question 11 - 2020 - Paper 1

To calculate the probability that Harry hits the bull's eye at least twice in three shots, we need to consider all the possible outcomes:

1. Hitting the bull's eye twice: (P(BB)) = (0.5 \times 0.5 \times 0.5 = 0.125 ) (Outcomes: BBN, BNB, NBB)
2. Hitting the bull's eye thrice: (P(BBB)) = (0.5 \times 0.5 \times 0.5 = 0.125 ) (Outcome: BBB)

Now, adding these probabilities:

$P(BB) + P(BBB) = 0.125 + 0.125 = 0.25$

Thus, the probability that Harry hits the bull's eye at least twice in his first three shots is (0.5).

In this scenario, where both Harry and Glenda have a 50% chance of hitting the bull's eye, the probability that the first shooter wins can be determined through an infinite geometric series:

$P = (0.5) + (0.5) imes (0.5)^2 + (0.5) imes (0.5)^4 + ...$

This series can be summed as:

1. The first term is (0.5).
2. The common ratio is (0.5^2 = 0.25).

Using the formula for the sum of an infinite geometric series, (S = \frac{a}{1 - r}), where (a) is the first term and (r) is the common ratio:

$S = \frac{0.5}{1 - 0.25} = \frac{0.5}{0.75} = \frac{2}{3}$

Therefore, the probability that the person who shoots first is the winner of the challenge is (\frac{2}{3}).