A. S= ( V,B), (V,E), (V,O), (M,B), (M,E), (M,O), (A,B), (A,E), (A,O)
B. Events are not likely. The quantities are not the same.
5.13 Are these characteristics of a student at your university mutually exclusive or not? Explain.
A. A=works 20 hours or more, B= majoring in accounting -not mutually exclusive
B. A= born in United States, B= born in Canada -mutually exclusive
C. A= owns a Toyota, B= owns a Honda -Not …show more content…
Let C be the event that a randomly chosen female aged 18-24 is Caucasian. Given P(S)=.246, P(C) =.830 and P(SC)= .232, find each probability and express the event in words. (Data are from Statistical Abstract of the United States, 2001.)
A. P(S).P(S)=1-.246. There is a 75.4% chance that a female aged 18-24 is a nonsmoker B. P(SC).-P(SC)=.246 + .830-.232=.844. There is an 84.4% chance that a female aged 18-24 is a smoker or is Caucasian C. P(SC).-P(SC)=.232/.830=.2795. Given that the female aged 18-24 is a Caucasian, there is a 27.95% chance that she smokes D. P(SC‘).-P(SC‘)=P(S)-P(SC)=.0824. Given that the female aged 18-24 is not Caucasian, there is an 8.42% chance that she smokes
5.23 Given P(A)= .40, P(B)=.50 and P(AB)=.05. (a). Find (AB). (b) In this problem, are A and B independent? Explain (a) P(AB)= P(AB)/P(B)=.05/.50=.10 (b) No because the multiplication of both is not equal to .5
5.33 | D | P | A | SUM | N | 13 | 6 | 6 | 25 | O | 58 | 30 | 21 | 109 | R | 8 | 7 | 7 | 22 | SUM | 79 | 43 | 34 | 156 |
a. P(D)=.5064 b.P(R) =.1410
c.P(DR)=.0513
d.P(DR)=.5962
e.P(RD)=.1013 …show more content…
(b) if the state allows any six-character mix (in any order) of 26 letters and 10 digits, how many unique plates are possible? (c) Why might some combinations of digits and letters be disallowed? * (d) Would the system described in (b) permit a unique license number for every car in the United States? For every car in the world? Explain your assumption. * (e) if the letters O and I are not used because they look too much like the numerals 0 and 1, how many different plates can be