Conditional probability question. I'm having trouble answering the question below:
You purchase a certain product. The manual states that the lifetime T of the product, defined as the amount of time (in years) the product works properly until it breaks down, satisfies
$P(T≥t)=e^{−t/5}$, for all $t≥0$.
For example, the probability that the product lasts more than (or equal to) 2 years is $P(T≥2)=e^{−2/5}=0.6703$. 
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?
It's a conditional probability question. I tried answering the following way, but my final answer doesn't seem to be correct.
$P(T≥2)=e^−2/5=0.6703$  =  Prob. that product lasts two or more years.
$P(T≥3)=e^−3/5= 0.5488$ = Prob. that product lasts three or more years.  
I figured that by subtracting the two $P(T≥2)-P(T≥3)$, I would be left with the probability that the product would break down within the third year.
Since we are calculating the prob that it breaks down within the third year, we are already assuming the condition that the product lasted two years. Thus the conditional should already included within the answer. Subtracting the two, I get the answer .1211, but answer seems to be .1813.
The answer can be found here:
https://www.probabilitycourse.com/chapter1/1_4_5_solved3.php
Thank you.
 A: You don't want to subtract; division is the operation you need.   As you said, it's a conditional probability problem.   So use the formula for that.
$$\begin{align}\mathsf P(X> 3\mid X>2) ~&=~ \dfrac{\mathsf P(X>3 \cap X>2)}{\mathsf P(X>2)} \\[1ex] ~&=~ \dfrac{\mathsf P(X>3)}{\mathsf P(X>2)}\end{align}$$
Then use $\mathsf P(X\leq 3\mid X>2) = 1-\mathsf P(X>3\mid X>2)$

Remark: The "the conditional is already included" bit you were thinking of may be that the event $X>3$ is a subset of the event $X>2$, and hence their conjunction is just the event $X>3$.

Addemdum: The result does demonstrates that $\mathsf P(X>3\mid X>2)=\mathsf P(X> 3-2)$, with which you may also have had that confused.   This is a consequence of a the memoryless property of exponential distributions.
A: By taking
$P(T≥2) - P(T≥3) = 0.1215$
You've actually worked out the chance of the product failing in the third year. $P(T\leq3\cap T\geq 2)$ The formula for the conditional probability of $A$ happening given $B$ has happened is 
$P(A|B)={\frac {P(A\cap B)}{P(B)}}$
$P(T\leq3|T\geq 2 )={\frac {P(T\leq3\cap T\geq 2)}{P(T\geq 2)}}$
$P(T\geq2)=0.6703$
$P(T\leq3\cap T\geq 2) = 0.1215$
$P(T\leq3|T\geq 2 ) = {\frac {0.1215}{0.6703}} = 0.1813$
A: Here's an analogy:
Pick three cards from the top of a standard deck. What's the probability the third card is a queen? Simply 4/52 as you don't know what the first or second cards were. However, suppose you know the first and second cards weren't queens? Then the probability increases to 4/50.
Consider also if the question had been "I purchase the product and use it for two years and it breaks down during that two-year period. What is the probability that it breaks down in the third year?"
In this case, the probability is zero that it breaks down in the third year (assuming it can only break once of course). In other words, what happens in the first couple years definitely has an effect on the probability of breakdown in the third year. If you know it didn't break down those first couple years, the probability it will break down the third year goes up.
