SOLUTION: If a man has a 0.7 chance of living in 20 years and his wife has 0.6. What is the chance that atleast one of them will live

Question 895784: If a man has a 0.7 chance of living in 20 years and his wife has 0.6. What is the chance that atleast one of them will live
Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
the man has a .7 probability to live 20 years.

his wife has a .6 probability to live 20 years.

the probability of at least one of them living 20 years is 1 minus the probability of neither of them living 20 years.

the probability that neither of them would live 20 years is equal to .3 *.4 = .12

the probability that at least one of them will live 20 years is therefore 1 - .12 = .88.

the breakdowns are as follows:

p(both will live 20 years) = .7 * .6 = .42

p(the man will live but the woman won't) = .7 * .4 = .28

p(the man won't but the woman will) = .3 * .6 = .18

p(both won't) = .12

add the probabilities up and they should equal 1.0

.42 + .28 + .18 + .12 = 1.0

probabilities check out ok so we're good.

solution is probability that at least one of them will live 20 years is equal to .88.

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