Answers
A) p(husband smoking)=.3
B) p(neither smokes)= .58
C) p(at least one smokes) = .42
D) p(at least one of them do not smoke) = .92
Explanation:
A--- the probability of the husband smoking is given.
B--- the probability of neither of the couple smoking is .58 probability of the wife smoking is .2, and the probability of both partners smoking is .08; so the probabilioty of only the wife who smokes but husband doesn't is .12; I got that from (probability of wife smokes) - (probability of wife and husdand smoke). The probability of the husband who doesn't who is .7; I got it from 1- (probabiloty of husband who smokes). So To find the probabilty of niether partner smoking, I have to find the differnce of husband who doesnt smoke from husband who doesn't smoke but his wife does. I got .58 from that. .7 - .12 =.58
C--- the probability of at leat one of them smoke is . 42. Because both are independent but not mutaullay explusive, I have to add the (probability of the hasband who smokes) and the (probability of the wife who smokes) and then minus the (probability of the husband and wife who smoke). I have to minus the )probability of both partners smoking because that probabilty is accounted for twice when you add the probabiloty of the husband smoking and the wife smoking. SO you have .3 + .2 - .08 = .42.
D--- The probability of at least one of the partner smokes is .92. You have to add the (probability of the couple where husband smokes but wife doesnt), the (probability of the couple where the wife smokes but the husband doesnt), and the (probability of the couple where neither of them smokes). So you would have..... .22+ .12 + .58 = .92.
This is a long explaination but I hope it helps.
