I purchase the product and use it for two years without any problems.
Painting A must be last, two possibilities are left for slot two, and just one is left for slot three.
The total number of possibilities when D is first is also 1 x 2 x 1 x 1 = 2.
Hard probability problems and counting problems, like most hard problems on the SAT, are not really “hard”.
The main thing to keep in mind is the technique that is discussed above for calculating possibilities in different contexts. We run a free online SAT prep seminar every few weeks.
Read more of his articles here, including How I Scored in the 99th Percentile and How to Effectively Study for the SAT.
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 $$P(T \geq t)=e^, \textrm t \geq 0.$$ For example, the probability that the product lasts more than (or equal to) $ years is $P(T \geq 2)=e^=0.6703$.
The only tricky part is this calculation will require two steps.
Imagine painting A is selected to go in slot one (one possibility for the desired outcomes).
The total desired outcome is all the possibilities when A is first (2) added to all the possibilities when D is first (also 2), which leaves a total desired outcomes of four.
With the total desired outcomes and the total possible outcomes found, the final step is to create a fraction with desired outcomes on the top and total outcomes on the bottom or 4/24 which reduces to 1/6.