Probability and Combinatorics

Question 1

The
probability of flipping a
fair coin once and getting
heads can be expressed as
1
2

Respectively, the
numerator and
denominator of this fraction
can be thought of as what?

Answer 1

**_# of possibilities that meet conditions_ 
# of equally likely possibilities**

Question 2

# _define_:
**trial** and **experiment**

Answer 2

Trial:
one
individual performance of a
random experiment

Experiment:
any procedure that can be
performed infinitely and has a
well-defined set of
possible outcomes
(basically a set of a number of trials)

Question 3

# _define_:
sample space

Answer 3

The

• *set** of
• *all possible outcomes**

• e.g.*
• In a six-sided di**, {1, 2, 3, 4, 5, 6}*

Question 4

What is the
name and
meaning of the
symbols below?

P(A)

Answer 4

name:
Probability function

meaning:
Probability of event A

Question 5

What is the
name and
meaning of the
symbols below?

P(A U B)

Answer 5

name:
Probability of events union

meaning:
Probability that of events A or B

Question 6

What is the
name and
meaning of the
symbols below?

P(A ∩ B)

Answer 6

name:
Probability of events intersection

meaning:
Probability that of events A and B

Question 7

What is the
name and
meaning of the
symbols below?

P(A | B)

Answer 7

name:
Conditional probability function

meaning:
Probability of event A given that B occurred

Question 8

If A and B are
mutually exclusive events,
then how would you
calculate
P(A U B) = _____?

Answer 8

P(A U B) = P(A) + P(B) − P(A ∩ B)

Add the
probabilities of each of them occurring
and
subtract the probability of them both occurring.

Subtract the overlap.

Question 9

What is the
addition rule for probability?

Answer 9

P(A U B) = P(A) + P(B) − P(A ∩ B)

For mutually exclusive events, add the probabilities of each of them occurring and subtract the probability of both occurring

• e.g.*
• To get to school, Harry can take the train, a car, or a broomstick. He has the same options when he leaves school. If he chooses his mode of transportation randomly to and from school, what is the probability that he takes a car at least once?*

P(C U C) = 1 + 1 − 1 = 5
3 3 9 9

Question 10

An average
deck of playing cards has
4 suits (H, C, S, D) and
13 ranks (2, 3, 4, 5, 6, 7, 8, 9, 10, J, Q, K, A),
totaling
52 cards.

If you
draw one card,
what is the probability that it is a
spades or a Jack?

Answer 10

P(S U J) = 16 = 4
52 13

P(S U J) = P(S) + P(J) − P(S ∩ J)

= 13 + 4 − 1
52 52 52

= 16
52

= 4
13

Question 11

# _define_:
independent event

Answer 11

One event

• *in no way affects**
• *what happens** in
• *other events**

Question 12

If you

• *add** the
• *probabilities** of
• *all possible**
• *independent events**, what do you get?

Answer 12

1

• Only independent events*
• e.g.*
• The probabilities of randomly picking each of the four cardinal directions is ¼.*
• Added together, this is ¼ + ¼ + ¼ + ¼ = 1*

Question 13

In craps, with
two fair six-sided dice,
how do you
calculate the probability of rolling
snake eyes?

Answer 13

• P(A* ∩ B) = P(A) • P(B),
• (given independent events A and B)*

• P(A* ∩ B) = P(A) • P(B)
• P(1₁* ∩ 1₂) = P(1₁) • P(1₂)

= 1/6 • 1/6

= 1/36

Question 14

On a multiple-choice exam question, there is
one correct answer, and the
probability of randomly selecting that answer is
x/y.

What is the
probability of randomly selecting an
incorrect answer?

Answer 14

1 − x/y

Question 15

How do you

• *calculate** the
• *probability** that a
• *fair coin** will land on
• *heads**
• *92 times** in a row?

Answer 15

P(H)⁹²

These are independent events, so you can
multiply the probabilities.

• P(H*) = ½
• P(H*)⁷ = (½)⁹²

= 1
4,951,760,157,141,521,099,596,496,896

≈ 2.01948 x 10⁻²⁸

Question 16

If you
randomly select
one of the 50 states,
50 times,
how can you
calculate the probability that you will select
Texas
at least once?

Answer 16

1 − (49/50)⁵⁰

P(Texas at least once)

= P(not only other states in 50 trials)

= 1 − P(only other states in 50 trials)

= 1 − (49/50)⁵⁰

≈ 0.63583

Question 17

William Shakespeare wrote

• *37 plays**, including
• Henry IV, Part 1* and
• Henry IV*, Part 2.

If you
randomly select two different Shakespearean plays to read,
how would you calculate the probability that you select those two plays
in that order?

Answer 17

P(A ∩ B) = P(A) • P(B | A)

Given
dependent events A & B,
t**he probability that A and B will occur is
equal to the
probability of A times the
probability of B given that A occurred.

P(1st_{Part 1} ∩ 2nd_{Part 2}) = P(1st_{Part 1}) • P(2nd_{Part 2} | 1st_{Part 1})

= 1 • 1
37 36

= 1
1,332

Question 18

Given fair dice,
how would you
calculate the
probability of rolling
two natural 20s on a d20 followed by
two 8s on a d8?

Answer 18

(1/20)² • (1/8)² = 1/256

These are independent events, so the respective probabilities can be multiplied

Question 19

Mathematically,
what does

n!

mean?

Answer 19

n factorial

or

n times
every natural number
less than n

or

n • (n − 1) • (n − 2) • … • 1

Question 20

What does

0!

mean?

Answer 20

1

0 factorial is
defined to be 1 so that the
formulas work

Question 21

# _define_:
permutation

Answer 21

An

• *arrangement** of
• *items** in a
• *particular order**

e.g.
ABCD and ABDC are
two permutations

Question 22

# _define_:
combination

Answer 22

An

• *arrangement** of
• *items** in which
• *order is irrelevant**

e.g.
ABCD and ABDC are
one combination

Question 23

What is the
permutations formula?

Answer 23

_nP_k = n!
(n − k)!

Question 24

What is the
combinations formula?

Answer 24

_nC_k = n!
k! • (n − k)!

Question 25

The Great Hall at Hogwarts has
four House tables:
Gryffindor, Hufflepuff, Ravenclaw, and Slytherin.

How would you
calculate the number of ways the
four House tables could be
arranged in a single-file row in the
Great Hall?

Answer 25

4!

Or four factorial, which is
4 • 3 • 2 • 1 = 24

It works like this . . .

• FOUR options for your first placement times
• THREE options for your second placement times
• TWO options for your third placement times
• ONE option for your final placement

or, using the permutations formula,

₄P₄ = 4!
(4 − 4)!

= 4!
0!

= 4!
1

= 4!

= 24

Question 26

The Great Hall at Hogwarts has
four House tables:
Gryffindor, Hufflepuff, Ravenclaw, and Slytherin.

How would you
calculate the number of ways the
four House tables could be
arranged in a single-file row in the
Great Hall if
Gryffindor and Slytherin cannot be next to one other?

Answer 26

4! − 2•3!

It works like this . . .

• 4! = 4•3•2•1, the number of permutations for placing 4 things in 4 spaces
• Must subtract all permutations with GS or SG next to one another
  
  • ​Treat GS as a unit
  • Permutations with H, R, and GS = 3! = 3•2•1 = 6
  • Must also account for SG
  • So must subtract 2•3!
• 4! − 2•3! = 12

Question 27

A deck of playing cards contains 52 unique cards.

What are

• *different ways** to
• *express** the number of ways a
• *deck of 52 playing cards** can be
• *shuffled**?

Answer 27

52!

or

52 • 51 • 50 • … • 2 • 1

or

₅₂P₅₂
(this is a
permutation because
order matters)

or

52!
(52 − 52)!

or

(or ≈ 8.06581752 × 10⁶⁷)

Question 28

A deck of playing cards contains
52 unique cards.

In five-card draw poker, a player is dealt a hand of
five cards that can be
reordered as the player wishes.

How do you calculate the number of
unique hands
that are possible?

Answer 28

₅₂C₅
(this is a
combination because
order does not matter)

or

52!
5! (52 − 5)!

(which is 2,598,960)

Question 29

There are 9 members of the United States Supreme Court: 1 Chief Justice and 8 Justices.

Before the Court hears oral argument, each of the
9 members shakes hands one time with
each other member.

How do you
calculate the number of
handshakes that take place before
oral argument?

Answer 29

₉C₂
(this is a
combination because
order does not matter:
Roberts/Kagan handshake =
Kagan/Roberts handshake)

or

9!
2! (9 − 2)!

(which is 36)

You divide by 2! so you don’t overcount:
if a Roberts/Kagan handshake occurs, a Kagan/Roberts handshake won’t occur.

Question 30

What is the
fundamental relationship
between
probability and
combinatorics?

Answer 30

probability • combinatorics

or

P(event happening once)
times the
(number of ways event can happen)

Question 31

A basketball player has a
70% free throw percentage.

How would you
calculate the probability of the
player making:

• *0 of next 4** free throws,
• *1 of next 4** free throws,
• *2 of next 4** free throws,
• *3 of next 4** free throws, and
• *4 of next 4** free throws.

Answer 31

The pattern:

P(event happening once)
times the
(number of ways event can happen)

0 of next 4 free throws:
(0.7⁰ • 0.3⁴) • ₄C₀
= 0.0081 • 1
= 0.0081

1 of next 4 free throws:
(0.7¹ • 0.3³) • ₄C₁
= 0.0189 • 4
= 0.0756

2 of next 4 free throws:
(0.7² • 0.3²) • ₄C₂
= 0.0441 • 6
= 0.2646

3 of next 4 free throws:
(0.7³ • 0.3¹) • ₄C₃
= 0.1029 • 4
= 0.4116

4 of next 4 free throws:
(0.7⁴ • 0.3⁰) • ₄C₄
= 0.2401 • 1
= 0.2401

The pattern:

P(event happening once)
times the
(number of ways event can happen)

Question 32

How does
combinatorics
relate to
Pascal’s triangle?

Answer 32

Basically,
it is Pascal’s triangle