Question 1 – Counting [5 points each]
a) How many odd integers are there from 1000 through 9999?
Solution:
First digit: 9 ways (exclude 0)
Second digit: 10 ways (as there are 10 possible
digits) Third digit: 10 ways (as there are 10 possible
digits) Fourth digit: 5 ways (oddnumber)
M.P: 9 *10*10*5 = 4500
Finally, 4500 possible odd numbers from 1000 to 9999
b) How many integers from 1000 through 9999 have distinct digits (i.e. 1234, 3592, 8710
butnot 1213, 2272, 8010)
Solution:
1st digit: 9 ways (exclude 0)
2nd digit: 9 ways (exclude 1st digit, include 0)
3rd: 8 ways (needs to be different from first and second digit)
4th digit: 7 ways (needs to be different from first, second and third digit)
multiplication rule:9 *9*8*7 = 4536
c) How many odd integers from 1000 through 9999 have distinct digits?
Solution:
5 choices for the last digit.
1st digit: 8 (exclude 0 and last digit)
8 choices for the 2nd digit.
7 choices for the 3rd digit
Odd numbers with distinct digits is (8 *8*7*5) = 2, 240
d) What is the probability that a randomly chosen four-digit integer has distinct digits and is
Odd?
probability that a randomly chosen four-digit integer has distinct digits 4536/9000=0.504
probability that a randomly chosen four-digit integer has distinct digits and isOdd?
2240/9000=0.2489
Question 2 – Joint and disjoint sets [10 points each]
A group of 8 people go to the movie together, and they sit next to each other.
a) Two of the people insist on sitting side-by-side. In how many ways can the eight be seated
together in a row?
Solution:
7! = 5040
5040*2=10 080 as Person A And Person B can sit either AB or BA
b) Other two people in the group do not like each other and do not want to sit side-by-side.
Now how many ways can the eight be seated together in a row?
Solution:
8!-7!*2=30240
Calculate all possible ways of arranging 8 people, and then using the difference rule we subtract the
unwanted cases.
Question 3 – Combinations [10 points]
16 people go on a trip.
They have a car with 5 seats, a car with 7 seats, and two motorbikes.
Considering that the owners of each car or motorbike wants to drive their own vehicle, how many
different ways can they sit in each vehicle?
Solution: Since 4 of 16 people are drivers of their own vehicle, we need to arrange (16-4) 12 people
1) 4 of 12 -> 5 seat car, 8 left 6
of 8 -> 7 seat car, 2 left
1 of 2 -> 2 seat motorbike, 1 left
1 of 1 -> 2 seat motorbike
2) 6 of 12 -> 7 seat car, 6 left
4 of 6 -> 5 seat car, 2 left
1 of 2 -> motorbike, 1 left
1 of 1 -> motorbike
(12!/4!8!)*(8!/6!2!)*(2!/1!
1!)=27720
Question 4 – Repetitions [5 points each]
A bakery produces six different kinds of pastry, one of which is éclairs. Assume there are at least 20 pastries
of each kind.
a) How many different selections of twenty pastries are there?
25!/20!5!=53130!
b) How many different selections of twenty pastries are there if at least three must be éclairs?
Solution (17!*22!)/(17!5!)=26334
c) How many different selections of twenty pastries contain at most two éclairs?
Solution: 53130-26334=26796
Question 5 – Conditional events [5 points each]
One urn contains 10 red balls and 25 green balls, and a second urn contains 22 red balls and
15 green balls. A ball is chosen as follows: First an urn is selected by tossing a loaded coin with
probability 0.4 of landing heads up and probability 0.6 of landing tails up. If the coin lands heads
up, the first urn is chosen; otherwise, the second urn is chosen. Then a ball is picked at random
from the chosen urn.
a) What is the probability that the chosen ball is green?
A=tossing coin; B=choosing a green ball
Solution: 𝑃(𝐵) = 0.4*25/35+0.06*15/37 = 0. 2857 + 0. 2432 = 0. 5289
b) If the chosen ball is green, what is the probability that it was picked from the first urn?
0.2857/ 0.5289=0.5402
Question 6 – Expected value [5 points each]
You play a game against opponents; each player bets $1 and the winner takes the whole prize. You
have a statistical edge, so your probability to win the game is 60% against any opponent.
a) What is the expected gain or loss in a single game?
expected value = (1-p)*loss + p*win = 0.6*1 - 0.4*1 =
0.2
b) What is the expected gain or loss in a 2-round tournament with 4 players, where the players
play a semi-final and the 2 winners play the final? Recall that the winner takes the whole prize of
$4 and you have a probability of 60% to win each game.
probability to win from semi is = 0.6*0.6 = 0.36
expected value = 0.36*3 - 0.64*1 = 0.44
c) What is the expected gain or loss in a 3-round tournament with 8 players, where the players
play a quarter-final, the 4 winners play a semi-final, and the 2 winners play the final for the
$8 prize?
expected value = 0.216*7 - 0.784*1 = 0.728
d) Is it more convenient to you to play 12 one-round tournaments with 2 players, 6 two-round
tournament with 4 players, or 4 three-round tournaments with 8 players?
Expected value for:
12 single games = 12*0.2 = 2.4
6 double games = 6*0.44 = 2.64
3 triple games = 3 * 0.728 = 2.184
the expected value is highest for 6 double games, but the probability to win a double match = 0.36 and it is
0.6 for single matches. Single match is more convenient.
e) In a three-round tournament, what is the expected value of number of games that you will
play?
100% chance to play 1st match
60 % to next
36 % play the last match
expected matches = 1 + 0.6 + 0.36 = 1.96