Question

An indicator light can be in one of three states: OFF, FLASHING and ON, with probabilities 1/2, 2/5 and 1/10 respectively. A test panel has five such lights whose states are mutually independent.
(a) What is the probability that three or more lights are OFF and at most one is ON?
(b) A second test panel with five such lights is observed independently. Let X be the total number of lights which are FLASHING among the two panels. Determine the probability mass function of X

Answer 1

Pr(OFF) = 1/2

Pr(Flashing) = 2/5

Pr(ON) = 1/10

(a) Pr(Three or more lights are OFF and atmost one is ON) = Pr(3 OFF, 1 ON , 1 Flashing)+ Pr(3 OFF, 0 ON, 2 Flashing) + Pr(4 OFF, 1 ON, 0 Flashing) + Pr(4 OFF, 0 ON, 1 Flashing) + Pr(5 OFF)

= ⁵C₃ (0.5)³* ²C₁ * (1/10) * (2/5) +  ⁵C₃ (0.5)³* ²C₂​* (2/5)² +  ⁵C₄ (0.5)⁴ *  (1/10) + ⁵C₄ (0.5)⁴*  (2/5)

+ ⁵C₅ (0.5)⁵

= 0.1 + 0.2 + 0.03125 + 0.125 + 0.03125 = 0.4875

(b) Here as both the lights have five panel.

so Here Pr(Flashing for each panel) = 2/5 = 0.4

so as we have x₁ from panel 1 and x₂ from panel 2

so here

x = x₁ + x₂

so here as two binomial distribution have same sample size and same probability then x have sample size = 5 +5 = 10 and p(success) = 0.40

p(x) = ¹⁰C_x (0.4)^x(0.6)^10-x