Maximize the probability of success to pass the exam As part of her computer science studies, Sam has to complete a mathematics examination. In the test, both simple and difficult tasks are set up. The probability of solving a simple problem is $ e $. A difficult task is usually solved with probability $ s $, where $ s <e $. Sam has to deal with three tasks to pass the exam. The tasks must be processed in the given order. Sam has the choice between easy-difficult-easy and difficult-easy-difficult. The test is passed if Sam successfully processes two consecutive tasks. It is assumed that the three tasks are processed successfully independently of each other. Which order should Sam choose to maximize the probability of success to pass the exam? 
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We have to calculate the possibilities $P(\text{easy-difficult-easy})$ and $P(\text{difficult-easy-difficult})$, or not? 
$P(\text{easy-difficult-easy})$ means that $P(\text{easy }\land \text{ difficult } \land \text{ easy})$ and since each of the is solved independently we get that $$P(\text{easy }\land \text{ difficult } \land \text{ easy})=P(\text{easy })\cdot P( \text{ difficult } )\cdot P(\text{ easy})=e\cdot s\cdot e=e^2\cdot s$$ 
Is that correct? 
Is the other possibility then the following? 
\begin{align*}P(\text{difficult-easy-difficult})&=P(\text{difficult }\land \text{easy }\land \text{ difficult}) \\ & =P( \text{difficult } )\cdot P(\text{ easy })\cdot P( \text{ difficult } ) \\ & =s\cdot e\cdot s=s^2\cdot e\end{align*} 
 A: As I understand the problem, success is defined by getting two questions in a row right.  Thus to succeed Sam has to get the middle one right, so I think intuition would suggest that Sam wants the middle one to be easy.  Let's check that.
Easy-Difficult-Easy:
Probability of getting all three right: $ese$
Probability of getting the first two right, third wrong:  $es(1-e)$
Probability of getting the first wrong, second two right:  $(1-e)se$.
So:  the success probability is $$ese+2es(1-e)=es+es(1-e)=es(2-e)$$
Difficult-Easy-Difficult
Probability of getting all three right: $ses$
Probability of getting the first two right, third wrong:  $se(1-s)$
Probability of getting the first wrong, second two right:  $(1-s)es$.
So:  the success probability is $$ses+2se(1-s)=se+se(1-s)=se(2-s)$$
Since $e>s\implies 2-s>2-e$ we see that the success probability is higher for Difficult-easy-difficult, confirming the intuition.
A: First, note that in order to correctly solve "two consecutive tasks", you must solve the middle task, and then either the first or last (or both) as well. In other words, you must pass the middle task and not fail both other tasks.
Probability of success if you choose hard-easy-hard: $e(1-(1-s)^2)$
$(1-s)$ represents failing the hard task, and you must not do this twice in a row. Failing it twice in a row occurs with probability $(1-s)^2$, so probability that you do not do this is $1-(1-s)^2$. Multiply by $e$ since you must pass the easy middle task. 
Probability of success if you choose easy-hard-easy: $s(1-(1-e)^2)$
$(1-e)$ represents failing the easy task, and you must not do this twice in a row. Failing it twice in a row occurs with probability $(1-e)^2$, so probability that you do not do this is $1-(1-e)^2$. Multiply by $s$ since you must pass the hard middle task.
Now, to the question: which is greater? Just choose a simple example where $s < e$. If $e=1$ and $s=0.5$, your probability of success with easy-hard-easy is $0.5$ and your probability of success with hard-easy-hard is $0.75$ Therefore, you should choose hard-easy-hard.
A: Interesting question!
If a pass for two consecutive right answers:
$1)$ Easy-hard-easy: 
$$P(SSF)+P(FSS)=es(1-e)+(1-e)se=2es(1-e).$$
$2)$ Hard-easy-hard:
$$P(SSF)+P(FSS)=se(1-s)+(1-s)es=2es(1-s)>2es(1-e).$$
If a pass for any two right answers:
$1)$ Easy-hard-easy:
$$P(SSF)+P(SFS)+P(FSS)=es(1-e)+e(1-s)e+(1-e)se=e^2(1-3s)+2es.$$
$2)$ Hard-easy-hard:
$$P(SSF)+P(SFS)+P(FSS)=se(1-s)+s(1-e)s+(1-s)es=s^2(1-3e)+2se<e^2(1-3s)+2es.$$
