Probability of engine failure A 2-engine airplane has 0.23 probability of an engine failure in general, 0.1 is probability of a failure in the right engine and 0.15 is the probability of a failure in the left engine.


*

*What is the probability of a failure in both engine?

*What is the probability of a failure in only one engine?


I solved it as follows:


*

*$P=0.1\cdot 0.15$

*$P=0.1\cdot(1-0.15)+0.15 \cdot(1-0.1)$


I didn’t develop intuition for probability yet and the fact that I didn’t use the probability of failure in general makes me think I got it all wrong.
 A: Your solution would be OK if the events involved were independent. But they aren't.
Call $A=$ failure in 1st engine and $B=$failure in second engine.
You must use this:
$$P(A)+P(B)=P(A\cup B)+P(A\cap B)$$
That is
$$0.1+0.15=0.23+P(A\cap B)$$
so the probability of a double failure is $0.02$.
A: 
2-engine airplane has 0.23 probability of an engine failure in general, 0.1 is probability of a failure in the right engine and 0.15 is the probability of a failure in the left engine.

"Engine failure in general" mean that "either the left or right engine fails".   So it's the union of the events.
You have: $\quad\mathsf P(L\cup R) = 0.23,\quad \mathsf P(R)=0.1,\quad \mathsf P(L)=0.15$  
So, clearly the engines don't fail independently.

1.What is the probability of a failure in both engine?

That is the probability of the intersection of the events.
Use: $\quad \mathsf P(L\cap R) = \mathsf P(L)+\mathsf P(R) - \mathsf P(L\cup R) $

2.What is the probability of a failure in only one engine?

That is the probability of the set difference of the union and the intersection of the events.  (Otherwise known as the "exclusive disjunction", or "xor").
Use: $\quad\mathsf P((L\cup R)\setminus (L\cap R)) = \mathsf P(L\cup R)-\mathsf P(L\cap R)$
A: The really problem is if it can suppose independence of the events.
In that case, the solution for the first question is
$$P(E_1 \cap E_2) = P(E_1)P(E_2)$$
Where $E_i$ means the engine $i$ fails.
And the solution for the second question,
$$P(E_1-E_2)=P(E_1\cap \bar E_2)\\
P(E_2-E_1)=P(E_2\cap \bar E_1)$$
Where $\bar A$ is the complement of the set $A$.
$E_1-E_2$ means that the event $E_1$ occurs but $E_2$ does not.
So, if only one engine fails at time, it equivalent to
$$P(\bar E_1\cap E_2)+P(E_2\cap \bar E_1)$$
And a theorem says that if two events are independents between them, then the the complement of one is independent with the other.
It means that, if $P(E_1 \cap E_2) = P(E_1)P(E_2)$ then $P(E_1 \cap \bar E_2) = P(E_1)P(\bar E_2)$.
Hence,
$$P(E_1\cap \bar E_2)+P(E_2\cap \bar E_1) = P(\bar E_1)P( E_2) + P(E_1)P(\bar E_2)$$
A: In general, the probability of a single failure of an engine is p.  The probability that one will fail on a twin-engine aircraft is 2p.  The probability that both will fail is p^2.  There are similar relationships for more engines.  2p^3, p^4, etc.
In this specific case, you have 0.1 in the right side and 0.15 on the left side, so you have 0.1 + 0.15 = 0.25 as the probability of either failing.  Not sure where your 0.23 came from.  For both to fail, you would get 0.1 x 0.15 = 0.015.
