What is an event?
What is an outcome of an event?
What is the probability of an outcome of an event?
Suppose you flip a fair coin. How many outcomes are possible?
What is an experiment?
What do call an outcome whose probability is 1?
What do call an outcome whose probability is 0?
How do you compute the probability of the outcome of an event
where all the outcomes are equally likely?
How do you compute the probability of the outcome of an event
where the outcomes are not equally likely?
What is an experiment’s sample space?
What is an elementary outcome?
Do all the elementary outcomes in a sample space have equal
probability?
What conditions must an elementary event satisfy?
What is the relation between an event and a sample space?
What are Kolmogorov’s axioms?
What is the relation between an event a sample space?
What is the relation between the probability of an event
occurring and the probability of an event not occurring?
What is the range of probabilities and an event?
What is the probability of the union of two events?
Let A and B be two events
What is the probability of A or B?
What is the probability of A and B?
Examples
What is the probability that a card selected from a deck is an
ace?
|S| = 52
|E| = 4
P(E) = |E|/|S| = 4/52 = 1/13
Let E be the event that the selected card is red. What is the
probability of E?
|E| = 26
|S| = 52
E = E1 U E2 U ... U E26
Ei Ω Ej = 0 when i ≠ j
P(E) = P(E1) + P(E2) …. + P(E26)
P(E) = 26/52 = 1/2
P(E) = P(E1) + P(E2) …. + P(E26)
What is the probability that a randomly selected integer chosen
from the first 100 positive integers is odd?
By saying that the integers were randomly selected the problem
is saying that all of the elementary outcomes are equally
likely.
|S| = 100
|E| = 50
P(E) = 50/100 = 1/2
S = {1,2,3,...,98,99,100|}
What is the probability that the sum of the numbers on two dice
is even when they are rolled?
(a,b) is an outcome of rolling two dice.
The sample space is
(1,1), (1,2)...(1,6)
(2,1), (2,2)...(2,6)
...
(6,1), (6,2)...(6,6)
Hence |S| = 36 or you can get the same result by using the
product rule,
|S| = 6 x 6 = 36
E1 and E2 where E1 is the event that the first die is even and
the second die is odd and E2 is the event that the first die is
odd and the second die is even.
P(E) = P(E1)
|E1| = 3 x 3
|E2| = 3 x 3
|E| = 18
P(E) = 18/36
+ P(E2) Why? E1 Ω E2 = 0
= 9
= 9
= 1/2
(1,2), (1,4), (1,6)
(3,2), (3,4), (3,6)
(5,2), (5,4), (5,6)
What is the probability that a card selected from a deck is an
ace or a heart?
A = you selected an ace
H = you selected a heart
What is A U H?
P(A U H) = P(A) + P(H) – P(A Ω H)
|A| = 4
|H| = 13
|A Ω H| = 1
P(A) = 4/52 = 1/13
P(H) = 13/52 = 1/4
P(A Ω H) = 1/52
P(A U H) = 4/52 + 13/52 – 1/52
= 16/52 = 4/13
As, Ac, Ad, Ah, Kh, Qh, Jh, 10h, 9h, 8h, 7h, 6h, 5h, 4h, 3h, 2h
Do you see that 16 cards satisfy the condition – Ace or Heart?
What is the probability that a card selected from a deck is an
ace and a heart?
|Ace and Heart| = 1
P(Ace and Heart) = 1/52
What is the probability that a card selected from a deck is an
ace and is red?
|A and R| = 2
P(A and R) = 2/52 = 1/26
What is the probability that when a coin is flipped two times in
a row, it lands heads up every time?
|S2| = 2 x 2 = 4
S2 = {(t,t), (t,h), (h,h), (h,t)}
E = event two heads turned up
|E| = 1
P(E) = 1/4
E = event three heads turned up
|S3| = 2 x 2 x 2 = 8
S3 = {(h,h,h), (h,h,t), (h,t,h), (h,t,t), (t,h,h), (t,h,t),
(t,t,h), (t,t,t)}
P(E) = 1/8
What is the probability that when a coin is flipped six times in
a row, it lands heads up every time?
|S6| = 2 x 2 x 2 x 2 x 2 x 2 = 26 = 64
|E| = 1 x 1 x 1 ... 1 = 1
P(E) = 1/64
What is the probability that a die never comes up an even number
when it is rolled six times?
(a,b,c,d,e,f)
E1 and E2 and ... and E6
|S| = 6 x 6 x ...6 = 66
|Ei| = 3
|E] = 3 x 3 x 3 ... 3 = 36
P(E) = 36/66 = (1/2)6 = 1/64
What is the probability of selecting none of the correct six
integers, where the order in which these integers are selected
does not matter, from the positive integers not exceeding a)40,
b)48 c)56 d)64?
|S| = C(40,6) = C(34 + 6, 6)
Ei = the event that ith ball was not one of the 6 correct balls.
|Ei| = 34
This is approach is wrong! Why?
|E| = C(34,6)
Why? The only way to guarantee that you select none of the
correct balls is to select the balls from a set that includes
none of the correct balls! How many balls are in the set of
balls that include none of the correct balls? That is, that
exclude all of the correct balls?
P(E) = C(34,6)/C(40,6)
Consider a simpler problem.
What is the probability of selecting none of the correct 3
integers, where the order in which these integers are selected
does not matter, from the positive integers not exceeding 7.
Suppose the correct numbers are 5,6,7
|C(7,3)| = 35
1,2,3; 1,2,4;
1,3,5; 1,3,6;
1,5,6; 1,5,7;
2,3,6; 2,3,7;
2,5,6; 2,5,7;
3,4,7; 3,5,6;
4,5,7; 4,6,7;
1,2,5;
1,3,7;
1,6,7;
2,4,5;
2,6,7;
3,5,7;
5,6,7;
1,2,6;
1,4,5;
2,3,4;
2,4,6;
3,4,5;
3,6,7;
1,2,7; 1,3,4;
1,4,6; 1,4,7;
2,3,5;
2,4,7;
3,4,6;
4,5,6;
|C(4,3)| = 4
1,2,3;
1,2,4;
1,3,4;
2,3,4;
The number of sets that exclude 5, 6, and 7
P(not 5,6 or 7) = 4/35
What is the probability of winning the lottery by selecting the
correct six integers, where the order in which these integers
are selected does not matter from the positive integers not
exceeding a)30 b)36 c)42, d)60?
In a super lottery, players win a fortune if they choose the
eight numbers selected by a computer from the positive integers
not exceeding 100. What is the probability that a player wins
the super lottery?
How many ways can a player select the 8 correct numbers?
|W| = 1
|S| = C(100,8)
P(W) = 1/C(100,8)
Suppose that 100 people enter a contest and that different
winners are selected at random for first, second, and third
prizes. What is the probability that Kumar, Janice, and Pedro
each win a prize if each has entered the contest?
Assume that no person wins more than 1 prize.
(1st,2nd,3rd)
100,99,98?
|S| = 100 x 99 x 98
|1st| = 3
|2nd| = 2
|3rd| = 1
|1st and 2nd and 3rd = 6
(J,P,K)(J,K,P)(P,J,K)(P,K,J)(K,J,P)(K,P,J)
|W| = 6
P(W) = 6/(100x99x98) = 6/(100x33x49)
Consider a simpler problem. Suppose that 5 people enter a
contest and that different winners are selected at random for
first and second prizes. What is the probability that Kumar and
Pedro each win a prize if each has entered the contest?
A,B,C,Kumar,Pedro
S = {A,B; A,C; A,Kumar; A,Pedro; B,A; B,C; B,Kumar; B,Pedro;
C,A; C,B; C,Kumar; C,Pedro; Kumar,A; Kumar,B; Kumar,C;
Kumar,Pedro; Pedro,A; Pedro,B; Pedro,C; Pedro,Kumar}
Why? You can choose the first place winner in 5 ways and the
second place winner in 4 ways. You can do both tasks in 5 x 4 =
20 ways
Probability(Kumar and Pedro) = 2/20 = 1/10
What is the probability that Abby, Barry, and Sylvia win the
first, second, and third prices, respectively, in a drawing if
200 people enter a contest and
a) No one can win more than one prize
b) Winning more than one prize is allowed
In roulette, a wheel with 40 numbers is spun. Of these, 18 are
red, and 18 are black. The two other numbers are 0 and 00. The
probability that when the wheel is spun it lands on any
particular number is 1/40
What is the probability that the wheel lands on a red number?
|R| = 18
|S| = 40
P(R) = 18/40 = 9/20
What is the probability that the wheel lands on a black number
twice in a row?
P(B and B) = ?
|S| = 40 x 40
An outcome looks like (n1,n2)
|B1B2| = 18 x 18 = 324
P(B1B2) = 324/(40x40)
What is the probability that the wheel lands on 0 or 00?
P(0 or 00) =
P(0) + P(00) – P(O and 00)
P(0) = 1/40
P(00)= 1/40
P(0 or 00) = 1/40 + 1/40 = 1/20
What is the probability that in five spins the wheel never lands
on either 0 or 00?
Not(0 or 00) = Not(0) and Not(00)
P(A) = 1 – P(Ac)?
P(A or Ac) = 1
A Ω Ac = the empty set
P(A or Ac) = P(A) + P(Ac) – P(A Ω Ac)
P(A or Ac) = P(A) + P(Ac)
P(Ac) = 1 – P(A)
If A is the event that 0 or 00 appeared. What is Ac? It is the
event where neither 0 or 00 appeared.
P(Ac) = 1- P(A)
P(neither 0 or 00) = 1 – 1/20 = 19/20
How many ways can 0 and 00 not turn up?
P(neither 0 or 00) = 38/40 = 19/20
What is the probability that the wheel lands on a number between
1 and 6, inclusive, on one spin, but does not land between them
on the next spin?