Independent solution

How to solve this Exponential Distribution question

Answer in brief

The equipment remains operational only when both independent exponential components exceed the five-year threshold. Multiplying their survival probabilities gives 0.054114, which rounds to choice B.

Setup

Setup

Let P and S be the two component lifetimes. Convert their respective means of 3 and 4 years to exponential rates.

P∼Exp⁡(1/3),S∼Exp⁡(1/4)P\sim\operatorname{Exp}(1/3),\qquad S\sim\operatorname{Exp}(1/4)
Pr⁡(P>t)=e−t/3,Pr⁡(S>t)=e−t/4\Pr(P>t)=e^{-t/3},\qquad \Pr(S>t)=e^{-t/4}

Model

Model

A series system survives the target duration precisely when both components survive it. Independence turns the joint survival probability into a product.

Pr⁡(min⁡(P,S)≥5)=Pr⁡(P≥5)Pr⁡(S≥5)\Pr(\min(P,S)\ge 5)=\Pr(P\ge 5)\Pr(S\ge 5)

Compute

Compute

Insert the two exponential survival probabilities and combine their exponents.

Pr⁡(min⁡(P,S)≥5)=e−5/3e−5/4\Pr(\min(P,S)\ge 5)=e^{-5/3}e^{-5/4}
e−5/3−5/4=e−35/12=0.0541137662…e^{-5/3-5/4}=e^{-35/12}=0.0541137662\ldots

Answer

Answer

The system survival probability is about 0.05 at the precision of the listed answers.

Pr⁡(min⁡(P,S)≥5)≈0.05(B)\boxed{\Pr(\min(P,S)\ge 5)\approx 0.05\quad\text{(B)}}