In an experiment a group of children each repeatedly throw a dart at a target - Edexcel - A-Level Maths Statistics - Question 3 - 2018 - Paper 2

To find P(H > 4), we can calculate it using the complement rule:

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

Where:

$P(H \leq 4) = P(H = 0) + P(H = 1) + P(H = 2) + P(H = 3) + P(H = 4)$

Using the binomial formula, we compute:

$P(H = k) = \binom{n}{k} p^k (1-p)^{n-k}$

Where: (p = 0.1), (n = 10). After computing, we find:

• P(H = 0) = 0.3487
• P(H = 1) = 0.3874
• P(H = 2) = 0.1937
• P(H = 3) = 0.0574
• P(H = 4) = 0.0123

Thus:

$P(H \leq 4) = 0.3487 + 0.3874 + 0.1937 + 0.0574 + 0.0123 = 0.9995$

So:

$P(H > 4) = 1 - 0.9995 = 0.0005$

Using Peta’s model, we know that F is the first success in a series of Bernoulli trials. Therefore:

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

For n = 5:

$P(F = 5) = (1 - 0.1)^{5-1} \cdot (0.1) = (0.9)^4 \cdot 0.1 = 0.06561$

To find α, we ensure the sum of probabilities equals 1:

$P(F = 1) + P(F = 2) + ... + P(F = 10) = 1$

Thus:

$0.01 + \left( \sum_{n=2}^{10} (0.01 + (n - 1) \cdot \alpha) \right) = 1$

Calculating:

$0.01 + 9 \cdot 0.01 + 0.02 + \left( 0.09 \cdot \alpha \right) = 1$

This gives:

$\alpha = 0.02$

Using Thomas’s model:

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

For n = 5:

$P(F = 5) = 0.01 + 4 \cdot 0.02 = 0.01 + 0.08 = 0.09$

Peta's model assumes the probability of hitting the target is constant for each throw, meaning every attempt is independent with the same success rate. In contrast, Thomas's model implies that the probability of hitting the target increases with each progressive throw, suggesting that children might improve or receive encouragement as they throw. Therefore, while Peta's model is static, Thomas's model is dynamic and relies on the number of attempts influencing success.