The problem consists of two parts. (a) A biased square spinner has the following probabilities: Score | 1 | 2 | 3 | 4 ------- | -------- | -------- | -------- | -------- Probability | 0.1 | 0.2 | 0.4 | 0.3 (i) Calculate the probability that one spin scores 2 or 3. (ii) Calculate the expected number of times to score 4 in 5000 spins. (iii) Calculate the probability of scoring 1 on the first spin and 4 on the second spin. (b) A bag contains 7 red discs and 5 blue discs. Two discs are drawn at random without replacement. Calculate the probability that at least one of the discs is red.

1. Problem Description

The problem consists of two parts.

(a) A biased square spinner has the following probabilities:

Score | 1 | 2 | 3 | 4

------- | -------- | -------- | -------- | --------

Probability | 0.1 | 0.2 | 0.4 | 0.3

(i) Calculate the probability that one spin scores 2 or

3. (ii) Calculate the expected number of times to score 4 in 5000 spins.

(iii) Calculate the probability of scoring 1 on the first spin and 4 on the second spin.

(b) A bag contains 7 red discs and 5 blue discs. Two discs are drawn at random without replacement.

Calculate the probability that at least one of the discs is red.

2. Solution Steps

(a)

(i) The probability of scoring 2 or 3 is the sum of their individual probabilities.

$P(2 \text{ or } 3) = P(2) + P(3)$

$P(2 \text{ or } 3) = 0.2 + 0.4 = 0.6$

(ii) The expected number of times to score 4 in 5000 spins is given by:

$E(\text{score 4}) = \text{Number of spins} \times P(4)$

$E(\text{score 4}) = 5000 \times 0.3 = 1500$

(iii) The probability of scoring 1 on the first spin and 4 on the second spin is given by the product of their individual probabilities. Since the spins are independent events, we multiply the probabilities.

$P(1 \text{ then } 4) = P(1) \times P(4)$

$P(1 \text{ then } 4) = 0.1 \times 0.3 = 0.03$

(b)

Total number of discs = 7 red + 5 blue =

1

2. We want to find the probability that at least one disc is red. It is easier to calculate the complement: the probability that both discs are blue.

$P(\text{both blue}) = P(\text{first blue}) \times P(\text{second blue | first blue})$

$P(\text{first blue}) = \frac{5}{12}$

If the first disc is blue, then there are 4 blue and 7 red, and 11 total.

$P(\text{second blue | first blue}) = \frac{4}{11}$

$P(\text{both blue}) = \frac{5}{12} \times \frac{4}{11} = \frac{20}{132} = \frac{5}{33}$

The probability of at least one red is:

$P(\text{at least one red}) = 1 - P(\text{both blue})$

$P(\text{at least one red}) = 1 - \frac{5}{33} = \frac{33}{33} - \frac{5}{33} = \frac{28}{33}$

3. Final Answer

(a) (i) 0.6

(ii) 1500

(iii) 0.03

(b) $\frac{28}{33}$