In an experiment a group of children each repeatedly throw a dart at a target - Edexcel - A-Level Maths Mechanics - Question 3 - 2018 - Paper 1

To find P(H > 4), we can use the complement rule: $P(H > 4) = 1 - P(H \, \leq \, 4)$

Using the binomial formula: $P(H = k) = {n \choose k} p^k (1 - p)^{n - k}$

Calculating for k = 0 to 4:

1. P(H = 0) = {10 \choose 0} (0.1)^0 (0.9)^{10} = 0.3487
2. P(H = 1) = {10 \choose 1} (0.1)^1 (0.9)^9 = 0.3874
3. P(H = 2) = {10 \choose 2} (0.1)^2 (0.9)^8 = 0.1937
4. P(H = 3) = {10 \choose 3} (0.1)^3 (0.9)^7 = 0.0574
5. P(H = 4) = {10 \choose 4} (0.1)^4 (0.9)^6 = 0.0128

Calculating P(H ≤ 4):

$P(H \leq 4) = P(H = 0) + P(H = 1) + P(H = 2) + P(H = 3) + P(H = 4) = 0.3487 + 0.3874 + 0.1937 + 0.0574 + 0.0128 = 1 - P(H > 4)$

Thus, $P(H > 4) = 1 - 0.9999 = 0.0001$

Using Peta’s assumptions, the random variable F follows a geometric distribution. To find P(F = n), we can use: $P(F = n) = (1 - p)^{n - 1} p$

For F = 5: $P(F = 5) = (0.9)^{4} (0.1) = 0.6561$

Using the model: $P(F = n) = 0.01 + (n - 1) × α$

For n = 1: $P(F = 1) = 0.01 + (1 - 1) × α = 0.01$

For n = 2: $P(F = 2) = 0.01 + (2 - 1) × α = 0.01 + α$

For n = 3: $P(F = 3) = 0.01 + (3 - 1) × α = 0.01 + 2α$

To satisfy the total probability: $P(F = 1) + P(F = 2) + P(F = 3) = 1$

Therefore, summing these: $0.01 + (0.01 + α) + (0.01 + 2α) = 1$

This simplifies to: $0.03 + 3α = 1$

Thus, α must equal: α = rac{1 - 0.03}{3} = 0.3233

Using the expression for Thomas' model: $P(F = 5) = 0.01 + (5 - 1) × α$

Substituting the value of α: $P(F = 5) = 0.01 + 4 × 0.02 = 0.09$

Peta’s model assumes that the probability of hitting the target remains constant at 0.1 for each child in each throw. This implies that every dart throw is independent.

Conversely, Thomas’ model suggests that the probability increases with each throw, implying that children are more likely to hit the target the more they throw. This introduces a dependency based on the number of attempts, which aligns more with a learning process rather than a fixed probability.