8. For brevity, let U = unemployed, E = employed, M = male, F = female. We're given that
P(M) = P(F) = 50/100 = 1/2
P(U) = 60/100 = 3/5
P(M | U) = 2/3
P(E) = 40/100 = 2/5
P(F | E) = 3/4
8a. This follows immediately from the given information. Specifically,
P(E) = 1 - P(U) = 1 - 3/5 = 2/5
8b. By definition of conditional probability,
P(A | B) = P(A and B) / P(B) ⇒ P(A and B) = P(A | B) P(B)
It follows that
P(M and U) = P(M | U) P(U) = 2/3 • 3/5 = 2/5
8c. Using Bayes' rule/the definition of conditional probability,
P(U | F) = P(U and F) / P(F) = P(F | U) P(U) / P(F)
Since F and M are mutually exclusive,
P(F | U) = 1 - P(M | U)
and so
P(U | F) = (1 - 2/3) • 3/5 / [(1 - 2/3) • 3/5 + 3/4 • 2/5] = 2/5
8d. Here we assume gender and employment status are independent, so for instance
P(F and E) = P(F) P(E)
We then have by the inclusion/exclusion principle that
P(F or U) = P(F) + P(U) - P(F and U) = P(F) + P(U) - P(F) P(U)
We also have by the law of total probability
P(F) = P(F and U) + P(F and E)
so
P(F or U) = P(F and U) + P(F and E) + P(U) - P(F) P(U)
By the assumed independence,
P(F or U) = P(F) P(U) + P(F) P(E) + P(U) - P(F) P(U)
P(F or U) = P(F) P(E) + P(U)
P(F or U) = 1/2 • 2/5 + 3/5 = 4/5
9.
a. This is mostly a matter of counting the ways a given type of stamp can fall out.
[tex]P(A) = \dfrac{\dbinom{20}3}{\dbinom{24}3} = \dfrac{285}{506}[/tex]
since there are 20 non-green stamps.
[tex]P(B) = \dfrac{\dbinom21 \dbinom{22}2}{\dbinom{24}3} = \dfrac{21}{92}[/tex]
since there are 2 red and unused stamps, 1 of which we want; the other 2 stamps come from the remaining 22 non-red-and-unused stamps.
[tex]P(A \cap B) = \dfrac{\dbinom21 \dbinom{18}2}{\dbinom{24}3} = \dfrac{153}{1012}[/tex]
since exactly 1 of the stamps must be red and unused, and the other 2 stamps that fall out can be neither green nor red and unused.
[tex]P(A \cup B) = P(A) + P(B) - P(A \cap B) = \dfrac{162}{253}[/tex]
which follows from the inclusion/exclusion principle.
b. There is a total of 10 used stamps, so the probability of at least 1 going missing is
[tex]P(C) = \dfrac{\dbinom{10}1\dbinom{14}2 + \dbinom{10}2\dbinom{14}1 + \dbinom{10}3}{\dbinom{24}3} = \dfrac{415}{506}[/tex]
By definition of conditional probability,
[tex]P(C \mid A) = \dfrac{P(C \cap A)}{P(A)}[/tex]
However, there are no used green stamps; any used stamp that goes missing must be red, blue or yellow. So the event A ∩ C is really just the event C, and
[tex]P(C \mid A) = P(C) = \dfrac{415}{506}[/tex]
c. A and C are independent if and only if
[tex]P(A \cap C) = P(A) P(C)[/tex]
We know
[tex]P(C \cap A) = P(C)[/tex]
so if A and C are independent, then
[tex]P(C) = P(A) P(C)[/tex]
but this would imply P(A) = 1, which is clearly not the case as we found in 9.a. So A and C are not independent.
