SOLUTION: If C is the event that a student will stay home to study and D is the event that she will instead go to a movie, P(C) = 0.70, and P(D) = 0.25, find (a) P(C'); (b) P(C or D).

Algebra ->  Probability-and-statistics -> SOLUTION: If C is the event that a student will stay home to study and D is the event that she will instead go to a movie, P(C) = 0.70, and P(D) = 0.25, find (a) P(C'); (b) P(C or D).       Log On


   

Question 953354: If C is the event that a student will stay home to study and D is the event that she will instead go to a movie, P(C) = 0.70, and P(D) = 0.25, find (a) P(C'); (b) P(C or D).
Answer by Theo(13342) About Me  (Show Source):
You can put this solution on YOUR website!
p(c) = .7
p(d) = .25

these events are mutually exclusive since she can't stay at home and study and go out to see a movie at the same time.

p(c and d) is therefore equal to 0.

p(c or d) = p(c) + p(d) - p(c and d) = .7 + .25 - 0 = .95.

p(c') is the probability that she won't stay home and study.

p(c') is equal to 1 - p(c) which is equal to 1 - .7 which is equal to .3

since all the probabilities added together must be equal to 1, she has:

p(c) = .7
p(d) = .25
p(c or d) = .95

p(c') = 1 - .7 = .3
p(d') = 1 - .25 = .75
p(c or d)' = 1 - .95 = .05

p(c or d)' means the probability that she will not stay home and study and she will not go out to a movies. in other words, she'll do something else.

the solution to your problem is:

p(c') = .3 and p(c or d) = .95.