Question

We have three light bulbs with lifetimes T1,T2,T3 distributed according to Exponential(λ1), Exponential(λ2), Exponential(λ3). In other word, for example bulb #1 will break at a random time T1, where the distribution of this time T1 is Exponential(λ1). The three bulbs break independently of each other. The three light bulbs are arranged in series, one after the other, along a circuit—this means that as soon as one or more light bulbs fail, the circuit will break. Let T be the lifetime of the circuit—that is, the time until the circuit breaks.(a) What is the CDF of T, the lifetime of the circuit?

Answer 1

Answer:

Step-by-step explanation:

Given that we  have three light bulbs with lifetimes T1,T2,T3 distributed according to Exponential(λ1), Exponential(λ2), Exponential(λ3).

Since connected in series, one after the other, along a circuit—this means that as soon as one or more light bulbs fail, the circuit will break.

Hence T  the lifetime of the circuit—that is, the time until the circuit breaks will be minimum (T1, T2, T3)

($P(T) = P(λ_1>T, λ_2>T, λ_3>T)\\=\pi P(λ_i>T)$

Since the three are independent

$P(T) = e^{-T(λ_1+λ_2+λ_3}$

CDF = $\int\limits^T_0 { e^{-T(λ_1+λ_2+λ_3}} \, dT \\=1- e^{-T(λ_1+λ_2+λ_3}$