Employment data at a large company reveal that 59% of the workers are married, that 20% are college graduates, and 1/6 of the college grads are married. what's the probability that a randomly chosen worker a) is neither married nor a college graduate? b) is married but not a college graduate? c) is married or a college graduate?

a) 113/300

b) 1/2

c) 187/300

Step-by-step explanation:

Lets call the events like follows:

a = married workers

b = graduate workers.

Then, a∧b = married and graduate workers.

We have:

p(a) = 59/100 (59%)

p(b) = 20/100 = 2/10 = 1/5 (20%)

p(a∧b) = 1/6.

So,

answer a) the probability to be neither married nor a college graduate is 1-p(aUb) = 1-(p(a)+P(b)-P(a∩B))=1-59/100-1/5+1/6= 113/300

answer b) the probability to be married but not a college graduate = p(a) ∩ (1-p(b)) = 59/100 x (1-20/100) = 1/2

answer c) the probability to be married or a college graduate p(a∪b)= p(a) ∪ p(b) -P(a∩B)= 59/100+1/5-1/6= 187/300

a) 0.25x times 3. (could be wrong on equation.)

b) 2.25 pound of dark chocolate. (can i eat all the chocolate? plz? anyway, again, could be wrong, but i hope im right. all the chocolate together should be 5.25 pounds)

have a good day!