The discrete random variable D has the following probability distribution $$ \begin{array}{|c|c|c|c|c|} \hline d & 10 & 20 & 30 & 40 & 50 \\ \hline P(D = d) & k & k & k & k \\ \hline \end{array} $$ where k is a constant - Edexcel - A-Level Maths Statistics - Question 4 - 2020 - Paper 1

To find ( P(D_1 + D_2 = 80) ), we note that the only combinations of ( D_1 ) and ( D_2 ) that will yield a sum of 80 are:

• ( D_1 = 30 ) and ( D_2 = 50 )
• ( D_1 = 40 ) and ( D_2 = 40 )
• ( D_1 = 50 ) and ( D_2 = 30 )

Each of these outcomes occurs with probabilities:

• ( P(D_1 = 30) \cdot P(D_2 = 50) = k \cdot k = k^2 )
• ( P(D_1 = 40) \cdot P(D_2 = 40) = k \cdot k = k^2 )
• ( P(D_1 = 50) \cdot P(D_2 = 30) = k \cdot k = k^2 )

Thus, we have:

$P(D_1 + D_2 = 80) = 3k^2$

Substituting value of k, we find:

$k = \frac{600}{137} \implies P(D_1 + D_2 = 80) = 3\left(\frac{600}{137}\right)^2$

Calculating yields:

( P(D_1 + D_2 = 80) \approx 0.0376 ) (to 3 significant figures).

The angles of quadrilateral Q are given as ( a, a + d, a + 2d, a + 3d ) where d is the common difference:

To find the smallest angle being more than 50\degree, we set:

1. ( a > 50 ) (Since if smallest angle is more than 50, this sets the least bound)
2. The sum of the angles must still equal 360\degree:

$a + (a + d) + (a + 2d) + (a + 3d) = 360$

This simplifies to:

$4a + 6d = 360 \quad \implies \quad a + \frac{3d}{2} = 90$

Thus, solving the inequalities:

• Substitute ( d = 10 ):

$10 > 50 \text{ giving us valid angles when } P(D = d) \text{ is evaluated}$

Establishing the probabilities for angles less than or equal to 50\degree results in:

To obtain exact probability, use:

$P(D = 20 \text{ or } 30) = P(D = 20) + P(D = 30) = 2k$

Thus, the overall probability:

• For common difference greater than 50\degree results in lower summed values. Hence final calculation yields an exact probability less than 1 thus ensuring it is more than 50\degree.