The discrete random variable Y has probability distribution | y | 1 | 2 | 3 | 4 | |---|-----|-----|-----|-----| | P(Y = y) | a | b | 0.3 | c | where a, b and c are constants - Edexcel - A-Level Maths Statistics - Question 3 - 2011 - Paper 2

To solve for P(3Y + 2 ≥ 8):

1. Rearranging the inequality:

   • Solve for Y: $3Y + 2 \geq 8 \Rightarrow 3Y \geq 6 \Rightarrow Y \geq 2$

2. Use probabilities for Y = 2, 3, and 4:

   • P(Y \geq 2) = P(Y = 2) + P(Y = 3) + P(Y = 4)
   • Calculate the total probability:
     • P(Y = 2) = b = 0.4
     • P(Y = 3) = 0.3
     • P(Y = 4) = c = 0.2
   • Thus: $P(3Y + 2 \geq 8) = b + 0.3 + c = 0.4 + 0.3 + 0.2 = 0.9$

Therefore, the answer is P(3Y + 2 ≥ 8) = 0.9.