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 complementary probability:

$P(H > 4) = 1 - P(H \leq 4)$

Using the binomial distribution: $P(H \leq 4) = \sum_{k=0}^4 P(H = k) = \sum_{k=0}^4 \binom{10}{k} (0.1)^k (0.9)^{10-k}$

Calculating this gives us:

$P(H \leq 4) = 1 - 0.01279 \approx 0.0128$

Thus, $P(H > 4) \approx 0.0128$

Using the geometric distribution since $F$ represents the number of trials until the first success:

$P(F = n) = (1 - p)^{n-1} p$ where $p = 0.1$.

Thus,

$P(F = 5) = (0.9)^{4} \cdot (0.1) \approx 0.06656.$

Therefore, $P(F = 5) \approx 0.0666.$

To find $\alpha$, we need to sum the probabilities up to 10.

$P(F = n) = 0.01 + (n - 1) \alpha$

Summing gives: $\sum_{n=1}^{10} P(F = n) = 1$

Thus, $10 \cdot 0.01 + \alpha \cdot \sum_{k=1}^{9} k = 1$

Where $ \sum_{k=1}^{9} k = \frac{9(9 + 1)}{2} = 45$$

Setting up the equation:

\Rightarrow \alpha = \frac{0.9}{45} = 0.02$$

Using Thomas' model for $P(F = n)$:

$P(F = n) = 0.01 + (n - 1) \cdot 0.02$

For $F = 5$:

$P(F = 5) = 0.01 + (5 - 1) \cdot 0.02$

This simplifies to: $P(F = 5) = 0.01 + 0.08 = 0.09$

Peta's model assumes the probability of hitting the target is constant for each throw, reflecting a typical binomial scenario.

In contrast, Thomas' model assumes that as trials increase, the probability of hitting the target increases, suggesting an improvement in skill or familiarity over time. This leads to a different interpretation of the success of hitting the target, emphasizing a learning effect through repeated attempts.