Question 895784
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.