Slide 2 Probability Presented by Tutorial Services The Math
Center Slide 3 Introduction to Probability Probability of an event.
Given an event E, the probability of the event, Pr(E), can be found
by: n(E) n(E) is the number of occurrences of the event. n(S) is
the number of elements in the sample space. n(S) =Pr(E) Slide 4
Empirical probability. Pr(event) = Number of occurrences of event
Number of total trials or observations Example: A coin was flipped
four times. It showed heads on the first flip, tails on the second,
heads on the third, and heads on the fourth. Find the empirical
probability of getting tails. Solution: Pr(Tails) = Number of
occurrences of tails Total number of trials = 1 4 Slide 5 Unions
and Intersections Pr (E and F both occur) = Pr (E F) Pr (E and F
both occur) = Pr (E F) Pr (E or F both occur) Pr (E or F both
occur) = Pr (E U F) Pr (E does not occur) Pr (E does not occur) =
Pr (E) E F E E S Slide 6 Solution: Solution: Pr (even)= Pr
(divisible by 3) = Pr (even & divisible by 3)=. Pr ( even or
divisible by 3) = _4_ + _3_ - _1_ = _6_ = _2_ 9 9 9 9 3 Inclusive
Events If E and F are two inclusive events, the probability that
one event or the other will occur is given by Pr (E or F) = Pr (E U
F) = Pr (E) + Pr (F) Pr (E F) Example: A card is drawn from a deck
of nine cards numbered 1 through 9. What is the probability that it
is even or divisible by 3? E F 3 9 1 9 4 9 Slide 7 Mutually
Exclusive Events Pr (E F) = 0, and Pr (E or F) = Pr (E U F) = Pr
(E) + Pr (F) For mutually exclusive events E 1, E 2, , E n Pr (E 1
U E 2 U U E n ) = Pr(E 1 ) + Pr(E 2 ) + Pr (E n ). EF S Slide 8
Solution: Solution: Let E be the event choosing first in class and
F be the event choosing second in class. Pr (E) = Pr (F) = Pr ( E U
F) = Pr (E) + Pr (F) = _98_ + _72_ = _170_ = _17_ 500 500 500 50
Example: Example: 500 students apply for a scholarship. 98 were
students who were first in their class and 72 were second in their
class. What is the probability that a student chosen at random is
first in their class or second in their class? 98 500 72 500 Slide
9 Conditional Probability Pr (A | B) = n (A B) n (B) The
Probability that A occurs, given that B occurs is given by, Pr (A |
B) = n (A B) n (B) Pr (A | B) = Pr (A B) Pr (B) Let A and B be
events in the sample space S with Pr (B) > 0. The conditional
probability that event A occurs, given that event B occurs is given
by, Pr (A | B) = Pr (A B) Pr (B) Slide 10 Solution: Solution:
Pr(sum is 9 sum >8) =Pr(sum is 9) = Pr(Sum > 8) = + + + = = 4
36 Pr(sum is 9 |sum >8) = 4/ 36 = 4 36 Example Example A pair of
6-sided die is rolled. What is the probability that the sum rolled
on the die is 9 given that the sum is greater than 8? 10/36 36. 10
= 4 = 2 5 Pr (A | B) = Pr (A B) Pr (B) 4 36 4 36 3 36 2 36 1 36 10
36 Slide 11 Product Rule for Events If A and B are probability
events, then the probability of the event A and B can be found by
using: i. Pr (A and B) = Pr (A B) = Pr (A) Pr (B|A) i. Pr (A and B)
= Pr (A B) = Pr (A) Pr (B|A) ii. Pr (A and B) = Pr (A B) = Pr (B)
Pr (A|B) ii. Pr (A and B) = Pr (A B) = Pr (B) Pr (A|B) If A and B
are independent events then: Pr(A and B) = Pr(A B) = Pr(A) Pr(B)
Slide 12 Solution: Solution: Pr(A) = Pr(pulling a face card) =
Pr(B) = Pr(tails on coin) = Pr(A B) = Pr(A) Pr(B) = 12 1 = 6 = 3
Example: Independent Events A card is pulled from a deck of 52
cards and a coin is tossed. Find the probability of pulling a face
card from the deck (Jack, Queen, and King) and getting a tails on
the coin. 5222652 12 52 1212 Slide 13 Probability Tree Draw a
probability tree when you have sequential occurrences. Draw a
probability tree when you have sequential occurrences. The
probability of the event occurring is the product of the
probabilities on the one particular branch. The probability of the
event occurring is the product of the probabilities on the one
particular branch. If an event can be described by two or more
paths through a probability tree, its probability is found by
adding the probabilities from the paths. If an event can be
described by two or more paths through a probability tree, its
probability is found by adding the probabilities from the paths.
Product of branch probabilities on path leading to F 1 through E 1
Sum of all branch products on paths leading to F1 =Pr(E1| F1) Slide
14 Example: The Queens jester had made a joke at the expense of the
Queen; usually punished by death, but the Queen liked this
particular jester. She was going to give him a chance to live. She
gave the jester 5 black balls, 5 white balls, and two urns and told
him, Put the balls in either urn as you feel. I will flip a coin
and pick an urn to pull a ball from. If the ball I pull is black,
then you will die. If its white, you will live. If the Jester puts
one white ball in the first urn and the rest in the second urn,
what is the probability that she chooses the first Urn given that
the ball he gets is white ? (1/2) (1) + (1/2) (4/9) 1/2 Urn 1 Urn 2
White Ball Black Ball 1 Solution: Pr(U1|W) = = 4949 (1/2) (1) 5959
9 13 Slide 15 Bayes Formula If a probability experiment has n
possible outcomes in its first stage given by E 1, E 2, , E n, and
if F 1 is an event in the second stage, then the probability that
event E 1 occurs in the first stage, given that F 1 has occurred in
the second, is Pr(E 1 | F 1 ) = Pr(E 1 ) Pr(F 1 | E 1 ) Pr(E1)
Pr(F1| E1) + Pr(E2) Pr(F1| E2) + Pr(En) Pr(F1 | En) Slide 16 Bayes
Formula The Queens jester had made a joke at the expense of the
Queen; usually punished by death, but the Queen liked this
particular jester. She was going to give him a chance to live. She
gave the jester 5 black balls, 5 white balls, and two urns and told
him, Put the balls in either urn as you feel. I will flip a coin
and pick an urn to pull a ball from. If the ball I pull is black,
then you will die. If its white, you will live. If the Jester puts
one white ball in the first urn and the rest in the second urn,
what is the probability that she chooses the first Urn given that
the ball he gets is white ? The Queens jester had made a joke at
the expense of the Queen; usually punished by death, but the Queen
liked this particular jester. She was going to give him a chance to
live. She gave the jester 5 black balls, 5 white balls, and two
urns and told him, Put the balls in either urn as you feel. I will
flip a coin and pick an urn to pull a ball from. If the ball I pull
is black, then you will die. If its white, you will live. If the
Jester puts one white ball in the first urn and the rest in the
second urn, what is the probability that she chooses the first Urn
given that the ball he gets is white ? Solution: Pr(U1) = 1/2
Pr(U2) = 1/2 Pr(W | U1) = 1 Pr(W | U2) = 4/9 Pr (U1) Pr(W |U1)_____
_ Pr (U1) Pr(W |U1) +Pr (U2) Pr(W |U2) = 1/2 = (1/2) (1) + (1/2)
(4/9) (1/2) (1) + (1/2) (4/9) 13/1813 (1/2) (1) = 9 Slide 17 Links
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