Download CmpE 343 Fall 2007 Problem Session#1 Solution Key Question1: In

CmpE 343
Fall 2007
Problem Session#1 Solution Key
Question1: In the “Sayısal Loto” game, a player selects 6 distinct numbers from 1 to 49 on a ticket
and “Milli Piyango” draws 6 distinct numbers at random from 1 to 49 in TV show “Sayısal Gece”. If all
the player’s numbers match the drawn ones then (s)he wins the jackpot and if 5 numbers match then
a smaller prize is won. Find the probability of each event.
Solution:
1
Pr(winning jackpot) = µ ¶ ≈ (0.7151)10−9
49
6
µ ¶µ ¶
6 43
1
5
Pr(winning smaller prize) = µ ¶ ≈ (0.1845)10−6
49
6
Question2: A lottery consists of 100 tickets, labeled 1, 2, . . . , 100, three of which are “winning numbers”. You buy four tickets. Calculate the probability that you have at least one winning ticket.
Solution:
Pr(at least one winning ticket) = 1 − Pr(no winning tickets)
96 95 94
= 1−
100 99 98
≈ 0.1164
Question3: A die is loaded in such a way that the probability that a 6 is thrown is 5 times that of any
other number, each of them being equally probable.
(a) By what factor is the probability of a total of 18 from 3 throws greater than that for an unloaded
die.
5
(b) Show that for the loaded die, the probability of obtaining a total of 6 from 4 throws is times
2
that of obtaining 5.
Solution:
(a)
Pr(
Pr(
µ ¶3
1
with loaded die)
2
= µ ¶3 = 27
with unloaded die)
1
6
(b)
µ
Pr(total of 6)
Pr(
=
Pr(total of 5)
) + Pr(
Pr(
)
)
1
10
¶4
µ
+6
µ ¶4
1
4
10
4
=
1
10
¶4
=
5
2
Question4: After having written n job application letters and addressing the n corresponding envelopes, your niece does you the kind favor of inserting one letter into each of the envelopes and mailing
them. However, she thought all the letters were the same, so that they were placed into the envelopes
in a completely random fashion.
(a) What is the probability that all companies get the correct letter?
(b) What is the probability that at least one is correct?
(c) What is the probability that none is correct?
Solution: Let Li be the event that letter i is in its correct envelope.
(a)
Pr(L1 ∩ L2 ∩ . . . ∩ Ln ) = Pr(L1 ) Pr(L2 |L1 ) Pr(L3 |L2 ∩ L1 ) . . .
1
1
1
...
=
n (n − 1) (n − 2)
1
=
n!
(b)
Pr(L1 ∪ L2 ∪ . . . ∪ Ln ) =
X
1≤i≤n
= 1−
X
Pr(Li ) −
Pr(Li ∩ Lj ) + . . .
1≤i<j≤n
1
1
1
+ − + ...
2! 3! 4!
(c)
Pr(none is correct) = 1 − Pr(L1 ∪ L2 ∪ . . . ∪ Ln )
1
1
1
= 1 − 1 + − + − ...
2! 3! 4!
1
≈
e
Question5: If n balls are individually dropped into r urns, where n ≥ r, such that the probability of
1
landing in any urn is . What is the probability that no urn is empty?
r
Solution: Let Ei be the event that urn i is empty.
Pr(some urn is empty) = Pr(E1 ∪ E2 ∪ . . . ∪ Er )
X
X
=
Pr(Ei ) −
Pr(Ei ∩ Ej ) + . . .
1≤i≤r
1≤i<j≤r
¶n
¶n
X µ
X µ
1
2
=
1−
−
1−
+ ...
r
r
1≤i≤r
1≤i<j≤r
µ ¶µ
¶n µ ¶ µ
¶n
r
2
r
1
−
1−
+ ...
=
1−
2
1
r
r
Pr(no urn is empty) = 1 − Pr(some urn is empty)
µ ¶µ
¶n µ ¶ µ
¶n
1
2
r
r
= 1−
1−
+
1−
− ...
1
r
2
r
µ ¶µ
¶n
X
i
i r
1−
=
(−1)
i
r
1≤i≤r