an article appears in a news paper stating that 40 of homeow
an article appears in a news paper stating that 40% of homeowners in a certain community have mortgages that are \"underwater\". In that same community 19% of the homeowners are unemployed or underemployed. Lasly, 3% of tr homeowners find themselves in both situations: homeowners with underwater mortgages and also unemployed or underemployed.
If a homeowner is chosen at random from this population, what is the probability that:
a) this person is unemployed/ underemployed or with an underwater mortgage.
b) this person is neither unemployed/ underemployed nor with an under water mortgage.
c) that this person is unemployed/underemployed but the mortgage is not under water.
also: If a homeowner that is neither unemployed/under employed is chosen at random, what is the probability that the mortgage is under water?
Thanks a million!!!
Solution
Given P(underwater) = 0.4
P(unemployed or underemployed) =0.19
P(homeowners with underwater mortgages and also unemployed or underemployed) =0.03
a) this person is unemployed/ underemployed or with an underwater mortgage.
P(unemployed/ underemployed or with an underwater mortgage)
=P(unemployed/ underemployed) + P(underwater mortgage) - P(both)
=0.19+0.4 -0.03
=0.56
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b) this person is neither unemployed/ underemployed nor with an under water mortgage.
P(neither unemployed/ underemployed nor with an under water mortgage)
= 1-0.56
=0.44
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c) that this person is unemployed/underemployed but the mortgage is not under water.
P(unemployed/underemployed but the mortgage is not under water)
=P(unemployed/underemployed) - P(both)
=0.19 - 0.03
=0.16
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also: If a homeowner that is neither unemployed/under employed is chosen at random, what is the probability that the mortgage is under water?
P(under water | neither unemployed/under employed)
=P(under water and neither unemployed/under employed)/P(neither unemployed/under employed)
= (0.4-0.03)/(1-0.19)
=0.4567901

