a. By definition of conditional probability,
P(C | D) = P(C and D) / P(D) ==> P(C and D) = 0.3
b. C and D are mutually exclusive if P(C and D) = 0, but this is clearly not the case, so no.
c. C and D are independent if P(C and D) = P(C) P(D). But P(C) P(D) = 0.2 ≠ 0.3, so no.
d. Using the inclusion/exclusion principle, we have
P(C or D) = P(C) + P(D) - P(C and D) ==> P(C or D) = 0.6
e. Using the definition of conditional probability again, we have
P(D | C) = P(C and D) / P(C) ==> P(D | C) = 0.75
Using probability concepts, we find that:
a) P(C AND D) = 0.3.
b) [tex]P(C \cap D) = 0.3 \neq 0[/tex], thus events C and D are not mutually exclusive.
c) Since [tex]P(C \cap D) \neq P(C)P(D)[/tex], they are not independent events.
d) P(C OR D) = 0.6.
e) P(D|C) = 0.75.
Item a:
By conditional probability, the relation is:
[tex]P(C|D) = \frac{P(C \cap D)}{P(D)}[/tex]
Considering [tex]P(C|D) = 0.6, P(D) = 0.5[/tex], we have that:
[tex]P(C \cap D) = P(C|D)P(D) = 0.6(0.5) = 0.3[/tex]
Thus:
P(C AND D) = 0.3.
Item b:
We have that [tex]P(C \cap D) = 0.3 \neq 0[/tex], thus events C and D are not mutually exclusive.
Item c:
[tex]P(C \cap D) = 0.3[/tex]
[tex]P(C)P(D) = 0.4(0.5) = 0.2[/tex]
Since [tex]P(C \cap D) \neq P(C)P(D)[/tex], they are not independent events.
Item d:
Using Venn Probabilities, we have that:
[tex]P(C \cup D) = P(C) + P(D) - P(C \cap D)[/tex]
Thus:
[tex]P(C \cup D) = 0.4 + 0.5 - 0.3 = 0.6[/tex]
Then P(C OR D) = 0.6.
Item e:
Using conditional probability:
[tex]P(D|C) = \frac{P(C \cap D)}{P(C)} = \frac{0.3}{0.4} = 0.75[/tex]
Thus P(D|C) = 0.75.
A similar problem is given at https://brainly.com/question/16723580
