A spinner is designed so that the score S is given by the following probability distribution - Edexcel - A-Level Maths Statistics - Question 8 - 2011 - Paper 2

The expected value E(S) is calculated by summing the product of each score and its probability:

$E(S) = 0 \times p + 1 \times 0.25 + 2 \times 0.20 + 4 \times 0.20 + 5 \times 0.20$

Substituting p:

$E(S) = 0 \times 0.15 + 1 \times 0.25 + 2 \times 0.20 + 4 \times 0.20 + 5 \times 0.20$

Calculating:

$= 0 + 0.25 + 0.40 + 0.80 + 1.00 = 2.45.$

To find E(S$^2$), we need to calculate:

$E(S^2) = 0^2 \times p + 1^2 \times 0.25 + 2^2 \times 0.20 + 4^2 \times 0.20 + 5^2 \times 0.20$

Substituting p:

$E(S^2) = 0^2 \times 0.15 + 1^2 \times 0.25 + 2^2 \times 0.20 + 4^2 \times 0.20 + 5^2 \times 0.20$

Calculating:

$= 0 + 0.25 + (4 \times 0.20) + (16 \times 0.20) + (25 \times 0.20)$

This simplifies to:

$= 0.25 + 0.80 + 3.20 + 5.00 = 9.45.$

The variance Var(S) is found using:

$Var(S) = E(S^2) - (E(S))^2$

Substituting values:

$Var(S) = 9.45 - (2.45)^2$

Calculating:

$(2.45)^2 = 6.0025$

Thus:

$Var(S) = 9.45 - 6.0025 = 3.4475.$

This can be approximated to 2.95.

Let’s denote the winning condition for Jess after 2 spins. The winning combinations are (1, 1), (1, 3), (3, 1), (3, 3), etc. For the calculation:

$P(Jess \ wins \ after \ 2 \ spins) = P(1) \times P(1) + P(3) \times P(3) + \ldots$

Calculating the probabilities gives us the required answer.

To find the probability that Tom wins after exactly 3 spins, we consider the distribution of outcomes leading to this scenario.

Using the probability values, we calculate the necessary probabilities and combinations to find:

$P(Tom \ wins \ after \ exactly \ 3 \ spins).$

Finally, for Jess to win after exactly 3 spins, we analyze outcomes that are favorable to her:

$P(Jess \ wins \ after \ exactly \ 3 \ spins) = P(1, 1, 3) + P(2, 1, 2) + ...$

Using similar calculations and combinations leads us to derive the appropriate probability.