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I have the problem and the answers as well that the prof has given the class, i'm trying to work backwards to figure out how he solved it. Would appreciate if anyone could show me how he got the numbers (answers are highlighted) as all he gave was a sort explanation.

  1. Suppose that the probability of a worker receiving the minimum wage is 10%. Of workers who receive the minimum wage, 50% are in low-income households. Of workers who receive more than the minimum wage, 20% are in low-income households.

(a) What is the probability that a random household has low income?

Composing the 2x2 table of joint probabilities and adding up the entries gives 0.05 + 0.18 = 0.23 or 23%

(b) If you are in a low-income household, what is the probability that you receive the minimum wage?

The conditional probability is 0.05/.23 = 21.74%.

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That problem is rather politically incorrect. It seems to be assuming a classical family with one worker per household. Without such an assumption, there's not enough information to answer it. (It also ignores household with no workers at all. There could be any number of these, and the question contains no information about their income.) –  joriki Oct 14 '12 at 19:33
    
@joriki you are right +1, but I presented the solution based on some common assumption, but there should not be any assumption you are right on that –  Diptarag Oct 14 '12 at 20:30
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1 Answer

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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%

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P(A ⋂ C) When you write this, what does the ⋂ denote in between the A and C? I also found that by writing a equation .5 = c / 0.1 and re-arranging to isolate c i got c = 0.05. Is this how you found the value 0.05? Thanks for the great answer as well! –  Alex Oct 14 '12 at 20:35
    
See by conditional probability P(C/A) = P(A ⋂ C) / P(A), and you know the value P(A) = 10% , P(C/A) = 50%, putting this you can find out the value of P(A ⋂ C) = 0.05 which implies probability that worker receives minimum wages(A) "and" their household is low income (C) = 0.05, similarly for other one –  Diptarag Oct 14 '12 at 20:41
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