“randomly”: given assignment
of some students to
certain slots, any of the remaining students is
equally likely to be assigned to any of the
remaining slots.
Question
: What is the probability that each
group includes a graduate student?

Multiplication Rule: Example 2
Denote the four graduate students by
1, 2, 3, 4
Define events
࠵?
1
= {
students 1 and 2 are in different groups
}
,
࠵?
2
= {
students 1, 2 and 3 are in different groups
}
,
࠵?
3
= {
students 1, 2, 3 and 4 are in different groups
}
.
We will use multiplication rule
࠵?
࠵?
3
= ࠵?
࠵?
1
∩ ࠵?
2
∩ ࠵?
3
= ࠵?
࠵?
1
࠵?
࠵?
2
࠵?
1
࠵?
࠵?
3
࠵?
1
∩ ࠵?
2
࠵?
࠵?
1
= 12/15
,
࠵?
࠵?
2
࠵?
1
= 8/14
,
࠵?
࠵?
3
࠵?
1
∩ ࠵?
2
= 4/13
.
So
࠵?
࠵?
3
=
12
15
∙
8
14
∙
4
13
≈ 0.14
.

The Monty Hall Problem
A prize is randomly
put behind one of the
three closed doors.
You point to one door.
A friend opens one of the remaining two
doors, after making sure that the prize is not
behind it.
Question
: Should you stick to your initial
choice, or switch to the other unopened
door?

The Monty Hall Problem
If sticking to the initial choice: the initial
choice determines whether you win or not,
Thus the winning probability is 1/3.
If switching
to the other unopened door:
Case 1: prize is behind the initial door, which
happens with probability 1/3
. You don’t win.
Case 2: prize is not behind the initial door, which
happens with probability 2/3. You win for sure.
So you should switch.

Content
Sets.
Probabilistic models.
Conditional probability.
Total Probability Theorem and Bayes’
Rule.
Independence.
Counting.

Total Probability Theorem
Let
࠵?
1
, ࠵?
2
, . . . , ࠵?
࠵?
be
disjoint
events that form a
partition
of the sample space. Assume
࠵?(࠵?
࠵?
) > 0
for
all
࠵?
. Then, for any event
࠵?
, we have
࠵?
࠵?
= ࠵?
࠵?
1
∩ ࠵?
+ ⋯ + ࠵?
࠵?
࠵?
∩ ࠵?
= ࠵?
࠵?
1
࠵?
࠵? ࠵?
1
+ ⋯ + ࠵?
࠵?
࠵?
࠵?
࠵? ࠵?
࠵?
Indeed,
࠵?
is the the disjoint union of
࠵?
1
∩ ࠵?
,
…
,
࠵?
࠵?
∩ ࠵?
.
The second equality is given by
࠵?
࠵?
࠵?
∩ ࠵?
= ࠵?
࠵?
࠵?
࠵?
࠵? ࠵?
࠵?
.

Example: chess tournament
Three types of players.
Type 1:
50%
Type 2:
25%
Type 3:
25%
You winning probability with these players:
Against type 1:
0.3
.
Against type 2:
0.4
.
Against type 3:
0.5
.
Now you play a game with a randomly chosen
player.
Question
: What’s your winning probability?

Example: chess tournament
࠵?
࠵?
: playing with an opponent of type
࠵?
࠵?
࠵?
1
= 0.5, ࠵?
࠵?
2
= 0.25, ࠵?(࠵?
3
) = 0.25
.
࠵?
: winning
࠵?
࠵? ࠵?
1
= 0.3,
࠵?
࠵? ࠵?
2
= 0.4,
࠵?(࠵?|࠵?
3
) = 0.5
The probability of
࠵?
:
࠵?
࠵?
= ࠵?
࠵?
1
࠵?
࠵? ࠵?
1
+ ࠵?
࠵?
2
࠵?
࠵? ࠵?
2
+ ࠵?(࠵?
3
)࠵?(࠵?|࠵?
3
)
= 0.50 × 0.3 + 0.25 × 0.4 + 0.25 × 0.5
= 0.375

Example: Four-Sided Die
Roll a fair 4-sided dice.
Rule: Roll once more if result is 1 or 2, otherwise
stop
Question
: What is the probability that the sum
total of your rolls is at least 4?

Example: Four-Sided Die
࠵?
࠵?
:
the result of first roll is
࠵?
࠵?
࠵?
࠵?
= 1/4
,
∀࠵? = 1,2,3,4.
࠵?:
the sum total is at least 4.
࠵?
࠵?
= σ
࠵?=1
4
࠵?
࠵?
࠵?
࠵?
࠵? ࠵?
࠵?
. Let’s calculate each
࠵?
࠵? ࠵?
࠵?
.
Given
࠵?
1
: the sum total will be
≥ 4
if the second roll
results in 3 or 4, which happens with probability
1/2
.
Thus
࠵?(࠵?|࠵?
1
) =
1
2
,
Similarly
࠵?(࠵?|࠵?
2
) =
3
4
.