Hi You can think like this -
Probability that a worker receiving min wage P(A) = 0.1
Probability that a worker receiving more than min wage P(B) = 0.9
Now probability that one household is low income provided the worker of that household receives min wages P(C/A) = 0.5 (C is the probability that the household is low income for minimum wages workers), again P(C/A) = P(A $\bigcap$ C) / P(A), so you can say P(A $\bigcap$ C) = 0.05 i.e total probability of the event : worker receives minimum wages "and" their household is low income = 5%
Similarly probability that one household is low income provided the worker of that household receives more than min wages P(D/B) = 0.2 (D is the probability that the household is low income for more than minimum wages workers), again P(D/B) = P(B $\bigcap$ D) / P(B), so you can say P(B $\bigcap$ D) = 0.18 i.e total probability of event : worker receives more than minimum wages "and" their household is low income = 18%
The event A:"worker receives minimum wages" and B:"worker receives more than minimum wages" are mutually exclusive, so you can come to conclusion any random house (not considering minimum wages) is low income household is : 0.18+0.05 = 0.23
And for second question "If you are in a low-income household, what is the probability that you receive the minimum wage?"
A: you are in a low-income household
B: you receive the minimum wage
By conditional probability
P(B/A) = P(A $\bigcap$ B)/P(A)
Remember we previously derived "probability of worker receives minimum wages "and" their household is low income = 5%" & probability that you are in a low-income household = 23%
So P(B/A) = 0.05/0.23 = 21.74%