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Here's the statement:

Quality control in a factory detects that 8% of the produced items are class $B$ quality only (i.e. not flawless). With a probability of 4% the production line functions suboptimally due to pollution, and with 50% probability the material used has quality issues. If pollution occurs, the items are class $B$ with a probability of 10%. Technicians also find that if there are fluctuations in the power supply then there is 10% probability that the product is class $B$. Based on these data, calculate the probability that power supply fluctuates.

I really have difficulty understanding the statement. As I see the problem: Let $P$ be the event "pollution", $Q$ the event a quality defect, $F$ the event "power supply fluctuate", and $B$ the event "$B$ quality".

I see the problem as follows. (see picture below).

So we just have
$$x=mathbb P(F)=1-0.04-0.5=0.46?$$
But I don't use $mathbb P(B)$. So maybe it should be $$mathbb P(B)=mathbb P(P)mathbb P(Bmid P)+mathbb P(Q)mathbb P(Bmid Q)+mathbb P(F)mathbb P(Bmid F)$$
which is equivalent to $$x=frac{0.01}{0.1}=0.1,$$
could it be possible?

I'll assume that the external factors (pollution, low quality, fluctuations) occur independently. I'll also assume that these are all the external factors. That is to say, if we find a Class $B$ item then we know that at least one of those external factors was present. Let $Phi$ denote the desired answer (the probability that fluctuations occur).

Let's use $Phi$ to compute the probability that an item isn't in Class $B$. Since we know that to be $.92$ we can then solve for $Phi$. We'll denote the three externals by $P,LQ, F$ and for any of those externals we let $X^c$ denote their complement. We'll go type by type.

Type $(P^c,LQ^c,F^c)$. The probability of this state occurring is $.96times .5times (1-Phi)$ and, of course, no item of this type is in Class $B$.

Type $(P^c,LQ^c,F)$ The probability of this state is $.96times .5times Phi$ and an item in this state is in Class $B$ with probability $.1$, hence it is not in class $B$ with probability $.9$

Type $(P^c,LQ,F^c)$ The probability of this state is $.96times .5times (1-Phi)$ and an item in this state is outside Class $B$ with probability $.9$

Type $(P,LQ^c,F^c)$ The probability of this state is $.04times .5times (1- Phi)$ and an item in this state is outside Class $B$ with probability $.5$

Type $(P^c,LQ,F)$ The probability of this state is $.96times .5times Phi$ and an item in this state is outside Class $B$ with probability $.9times .9=.81$

Type $(P,LQ^c,F)$ The probability of this state is $.04times .5times Phi$ and an item in this state is outside Class $B$ with probability $.5times .9=.45$

Type $(P,LQ,F^c)$ The probability of this state is $.04times .5times (1-Phi)$ and an item in this state is outside Class $B$ with probability $.5times .9=.45$

Type $(P,LQ,F)$ The probability of this state is $.04times .5times Phi$ and an item in this state is outside Class $B$ with probability $.5times .9times .9=.405$

Thus we get the (somewhat unpleasant) linear equation:

$$.48times (1-Phi)times 1+ .48times Phitimes .9+ .48times (1-Phi)times .9+ .02times (1-Phi)times .5+ .48times Phitimes .81+cdots $$
$$cdots +.02times Phitimes .45 +.02times (1- Phi)times .45+ .02times Phitimes .405=.92$$

And if no errors have been made (a big if, I grant) this comes to $$Phi=frac {110}{931}approx .11815$$

Note: the above is badly error prone so I strongly advise checking each stage carefully.

A simpler version: Let's only consider the three events $E_P,E_{LQ},E_F$ where, for an external variable $X$, $E_X$ denotes the event "$X$ is present and causes the item to be in Class $B$." Then we easily compute $$P(E_P)=.04times .5=.02quad P(E_{LQ})=.5times .1=.05quad P(E_F)=Phitimes .1$$ Then the probability that an item is not in class $B$ is $$.92=(1-.02)(1-.05)(1-.1times Phi)$$

Which again implies $$Phi=boxed {frac {110}{931}}$$