You purchase a certain product. The manual states that the lifetime TT of the product, defined as the amount of time (in years) the product works properly until it breaks down, satisfies P(T≥t)=e−t5, for all t≥0. For example, the probability that the product lasts more than (or equal to) 22 years is P(T≥2)=e−25=0.6703P(T≥2)=e−25=0.6703. I purchase the product and use it for two years without any problems. What is the probability that it breaks down in the third year?

You purchase a certain product. The manual states that the lifetime T$T$ of the product, defined as the amount of time (in years) the product works properly until it breaks down, satisfies

P(T≥t)=e−t5, for all t≥0.$$$$

For example, the probability that the product lasts more than (or equal to) 2$2$ years is P(T≥2)=e−25=0.6703$P(T\ge 2)=e^{-\frac{2}{5}}=0.6703$. I purchase the product and use it for two years without any problems. What is the probability that it breaks down in the third year?

• Solution
  

  • Let A$A$ be the event that a purchased product breaks down in the third year. Also, let B$B$ be the event that a purchased product does not break down in the first two years. We are interested in P(A|B)$P(A|B)$. We have
    
    

    P(B)$P(B)$	=P(T≥2)$=P(T\ge 2)$
    	=e−25$=e^{-\frac{2}{5}}$.

    
    We also have
    

    P(A)$P(A)$	=P(2≤T≤3)$=P(2\le T\le 3)$
    	=P(T≥2)−P(T≥3)$=P(T\ge 2)-P(T\ge 3)$
    	=e−25−e−35$=e^{-\frac{2}{5}}-e^{-\frac{3}{5}}$.

    
    Finally, since A⊂B$A\subset B$, we have A∩B=A$A\cap B=A$. Therefore,
    
    

    P(A|B)$P(A|B)$	=P(A∩B)P(B)$=\frac{P(A\cap B)}{P(B)}$
    	=P(A)P(B)$=\frac{P(A)}{P(B)}$
    	=e−25−e−35e−25$=\frac{e^{-\frac{2}{5}}-e^{-\frac{3}{5}}}{e^{-\frac{2}{5}}}$
    	=0.1813$<b>=</b>\mathrm{<b>0.1813</b><br><br>}$