The discrete random variable D has the following probability distribution | d | 10 | 20 | 30 | 40 | 50 | |----|----|----|----|----|----| | P(D = d) | k | k | k | k | k | where k is a constant - Edexcel - A-Level Maths Statistics - Question 4 - 2020 - Paper 1

Using the independence of D1 and D2:

We consider the pairs that can sum to 80:

• (30, 50)
• (40, 40)
• (50, 30)

Calculating the probabilities:

1. For (30, 50): $P(D_1 = 30) imes P(D_2 = 50) = k imes k = k^2$

2. For (40, 40): $P(D_1 = 40) imes P(D_2 = 40) = k imes k = k^2$

3. For (50, 30): $P(D_1 = 50) imes P(D_2 = 30) = k imes k = k^2$

Total Probability:

P(D_1 + D_2 = 80) &= P(30, 50) + P(50, 30) + P(40, 40) \\ &= 3k^2 \\ &= 3\left(\frac{600}{137}\right)^2\\ &= 3\left(\frac{360000}{18769}\right) \\ &\approx 0.0376.\end{aligned}$$ Thus, the probability is approximately 0.0376.

For the angles of quadrilateral Q: Let the angles be: $a$, $a + d$, $a + 2d$, $a + 3d$.

Where the sum of angles in a quadrilateral is 360°:

4a + 6d &= 360 \\ \Rightarrow 2a + 3d &= 180 \\ \Rightarrow a + d &> 50° \\ \\ a + d &> 50 \\ \Rightarrow \text{Smallest angle is } a = 50°\text{ gives us cases to consider} \\ \Rightarrow 10° \leq 75° \text{ as cases} \\ P(D = 10 ext{ or } 20)\text{ gives } = \frac{10}{137} + \frac{20}{137} \\ \text{Thus: } P \text{(Smallest angle > 50°)} \text{ is 0.657 or } \approx 0.657.\end{align*}$$