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Math 463/ Homework #2 - Solutions
- How many 5-digit numbers can be formed with the integers 1, 2 ,... , 9 if no digit can appear exactly once? (For instance 43443 is OK, while 43413 is not allowed.)
SOLUTION: There could be no more than two different digits in the number. First count the numbers formed out of exactly one digit, such as 77777. There are 9 of them. Next count the numbers made of two digits, call them digit a and digit b, where a appears twice, and b appears three times in a number (for example abbab). There are 9 × 8 = 72 choices for a and b, and (^) 2!3!5! = 10 ways to arrange two a and three b in a string of five digits. Thus the answer is 9 + 72 × 10 = 729.
- We shuffle a deck of six cards 1 , 2 , 3 , 4 , 5 , and 6 so that each of the 6! possible configurations (orderings) has equal probability of (^) 6!^1. Let A be the event that card 1 is among the top three cards in the deck, and let B be the event that card 5 ends up being second from the top. Find the probability of the event A ∪ B.
SOLUTION: First, we compute P (A ∩ B). There is only one way to place card 5 : it will take the second spot from the top. There are 2 spots left for card 1 : it can be placed either first or third. There are 4! ways to arrange the rest of the cards. Thus |A ∩ B| = 2 × 4! = 48 and P (A ∩ B) = (^48) 6! = 151. Trivially, P (A) = 12 , and P (B) = 16. Thus, by the inclusion-exclusion theorem,
P (A ∪ B) = P (A) + P (B) − P (A ∩ B) =
- Simplify (E ∪ F ) ∩ (E ∪ F ), and (E ∪ F ) ∩ (E ∪ F ) ∩ (E ∪ F )
SOLUTION:
(E ∪F )∩(E ∪F ) =
E ∩(E ∪F )
F ∩(E ∪F )
= E ∪(E ∩F )∪(F ∩E)∪(F ∩F ) = E
and therefore
(E ∪ F ) ∩ (E ∪ F ) ∩ (E ∪ F ) = F ∩ (E ∪ F ) = (F ∩ E) ∪ (F ∩ F ) = E ∩ F
- Consider events E, F ,and G. Find the expressions in E, F and G for the following events.
(a) At least one of the three events occurs. (b) At most one of the three events occurs.
(c) Exactly two of them occur. (d) At most two of the three events occur. (e) All three events occur. (f ) None of the three events occurs. (g) At most three of the events occur. (h) E or F , but not G occur. (i) Both E and F , but not G occur. (j) Exactly one of the three events occurs.
Example: The event that “only G occurs” is expressed as E ∩ F ∩ G.
SOLUTION:
(a) E ∪ F ∪ G (b) (E ∩ F ∩ G) ∪ (E ∩ F ∩ G) ∪ (E ∩ F ∩ G) ∪ (E ∩ F ∩ G) (c) (E ∩ F ∩ G) ∪ (E ∩ F ∩ G) ∪ (E ∩ F ∩ G) (d) E ∩ F ∩ G = E ∪ F ∪ G (e) E ∩ F ∩ G (f ) E ∪ F ∪ G = E ∩ F ∩ G (g) S (h) (E ∪ F ) ∩ G (i) E ∩ F ∩ G (j) (E ∩ F ∩ G) ∪ (E ∩ F ∩ G) ∪ (E ∩ F ∩ G)
- Prove that
P (E∪F ∪G) = P (E)+P (F )+P (G)−P (E∩F ∩G)−P (E∩F ∩G)−P (E∩F ∩G)− 2 P (E∩F ∩G)
SOLUTION: The inclusion-exclusion rule implies
P (E ∪F ∪G) = P (E)+P (F )+P (G)−P (E ∩F )−P (E ∩G)−P (F ∩G)+P (E ∩F ∩G),
where, after plugging in
P (E ∩ F ) = P (E ∩ F ∩ G) + P (E ∩ F ∩ G),
P (E ∩ G) = P (E ∩ F ∩ G) + P (E ∩ F ∩ G)
- Use induction to generalize Bonferroni’s inequality to n events:
P (E 1 ∩ E 2 ∩... ∩ En) ≥ P (E 1 ) + P (E 2 ) + · · · + P (En) − (n − 1)
SOLUTION: We know this is true for n = 2:
P (E 1 ∩ E 2 ) ≥ P (E 1 ) + P (E 2 ) − 1
For n ≥ 2, suppose we established the Bonferroni’s inequality for any n events:
P (E 1 ∩ E 2 ∩... ∩ En) ≥ P (E 1 ) + P (E 2 ) + · · · + P (En) − (n − 1)
We need to prove the inequality for any n + 1 events:
P (E 1 ∩ E 2 ∩... ∩ En ∩ En+1) ≥ P (E 1 ) + P (E 2 ) + · · · + P (En) + P (En+1) − n
We use the Bonferroni’s inequality for any n events, E 1 , E 2 ,... , En− 1 and En ∩ En+1, to get
P (E 1 ∩E 2 ∩.. .∩En− 1 ∩(En∩En+1)) ≥ P (E 1 )+P (E 2 )+· · ·+P (En− 1 )+P (En∩En+1)−(n−1),
where P (En ∩ En+1) ≥ P (En) + P (En+1) − 1, and therefore
P (E 1 ∩E 2 ∩.. .∩En∩En+1) ≥ P (E 1 )+P (E 2 )+· · ·+P (En− 1 )+P (En)+P (En+1)− 1 −(n−1)
- Show that the probability that exactly one of the two events, E or F , occurs is
P (E) + P (F ) − 2 P (E ∩ F )
SOLUTION: (E ∩ F ) ∪ (E ∩ F ) is the event that exactly one of the two occurs. The inclusion-exclusion formula implies
P
(E ∩ F ) ∪ (E ∩ F )
= P (E ∩ F ) + P (E ∩ F ) = P (E) + P (F ) − 2 P (E ∩ F )
as P (E ∩ F ) = P (E) − P (E ∩ F ) and P (E ∩ F ) = P (F ) − P (E ∩ F )
- Prove that if E 1 , E 2 ,... , En are independent events, then
P (E 1 ∪ E 2 ∪... ∪ En) = 1 −
∏^ n
i=
1 − P (Ei)
SOLUTION:
P (E 1 ∪ E 2 ∪... ∪ En) = 1 − P ((E 1 ∪ E 2 ∪ · · · ∪ En)c) = 1 − P (E 1 ∩ E 2 ∩... ∩ En) by De Morgan’s law. Hence, by independence,
P (E 1 ∪ E 2 ∪ · · · ∪ En) = 1 − P (E 1 ∩ E 2 ∩... ∩ En) = 1 −
∏^ n
i=
P (Ei) = 1 −
∏^ n
i=
(1 − P (Ei))
- An urn contains n green and m black marbles of the same size, shape, and weight. The marbles are drawn from the urn at random one at a time so that no preference is given to one marble over others, until only those of the same color are left. Show that with probability (^) n+nm they are all green.
SOLUTION: The outcome of the experiment will not change if you pull all but one last marble from the urn, and then check its color. Now, removing n + m − 1 marbles from the urn, and checking the color of the last marble is no different from selecting one marble and checking its color. Both are equivalent to separating marbles into two groups, one of size one, and the other of size n + m − 1. Thus the marble is green with probability (^) n+nm.
