Download Review for MAT 114 Exam 2-PDF

Math 114 – Review for Exam #2
Sections 7.1,7.2,7.3,7.4,7.5,8.2,8.3,& 8.5
For the exam you should know how to do the following problems:
(1) Define a Sample Space and determine the size of a sample space.
The Sample Space represents all the possible outcomes that can occur for a given
experiment. The size of a Sample Space is denoted as c(S).
Examples:
A) Define the Sample Space for flipping a coin. What is the size of this Sample
Space?
S = {T,H}
c(S) = 2
B) Define the Sample Space for flipping a coin 3 times. What is the size of this
Sample Space?
S = {HHH,HHT,HTH,THH,TTH,THT,HTT,TTT}
The size of the sample space can be found by using the multiplication principle.
c(S) = 2 x 2 x 2 = 8
= # of choices for first flip of coin X # of choices for second flip of coin X #
of choices for third flip of coin.
You should also know how to construct a tree diagram for this sample space.
C) Define the Sample Space for throwing a pair of dice. What is the size of this
Sample Space?
Size of the Sample Space = 6 x 6 = 36
= # of choices for first die X # of choices for second die
S = {(1,1) (1,2) (1,3) (1,4) (1,5) (1,6) (2,1) (2,2) (2,3) (2,4) (2,5) (2,6) (3,1) (3,2) (3,3)
(3,4) (3,5) (3,6) (4,1) (4,2) (4,3) (4,4) (4,5) (4,6) (5,1) (5,2) (5,3) (5,4) (5,5) (5,6) (6,1)
(6,2) (6,3) (6,4) (6,5) (6,6)}
You should also know how to construct a tree diagram for this sample space.
(2) Assign Valid Probabilities to outcomes in a sample space.
Remember – (1) Probability can not be negative, (2) Probabilities can not be greater than
1, and (3) The sum of all the probabilities in a given sample space equals 1.
For a sample space with equally likely outcomes, the probability that a desired outcome
occurs:
P(of desired outcome) = The number of ways desired outcome can occur
The total number of possible outcomes
1
Examples:
A) A die is tossed, find the probability that a 6 is rolled.
P(6) = Number of ways a six can occur
Number of total Outcomes (c(S))
P(6) = 1/6
B) A pair of dice is rolled find the probability that the sum of the roll is 5.
First find the number of total outcomes (c(S))
= # of choices for first die x # of choices for second die
= 6 x 6 = 36
Then find the number of ways the dice rolled can add up to 5
{(1,4) (2,3) (3,2) (4,1)} = 4 ways
P ( sum5) =
4 1
=
36 9
C) A hat contains 10 red marbles, 5 blue marbles, 4 pink marbles, and 1 black marble
– what is the probability you pick a red marble out of the hat?
What is the number of red marbles? 10
What is the total number of marbles to pick from? 20
P (red ) =
10 1
=
20 2
(3) Construct a Probability Model
A coin is tossed. The coin is weighted so that heads (H) is 6 times more likely to
occur than tails (T). Construct a probability model for this experiment.
You know that there are only two outcomes when you flip a coin heads (H) or tails
(T) so the sum of the probability of getting a head and the probability of getting a tail
is equal to 1:
P(H) + P(T) = 1
You also know that the probability of getting a head (H) is 6 times more likely than
the probability of getting a tail (T) so let P(H) = 6x and P(T) = x, then
6x + x = 1
x = 1/7
then, P(H) = 6/7 and P(T) = 1/7
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(4) Find the probability of mutually exclusive events occurring
Mutually Exclusive events are events that can not occur at the same time. For instance
when you roll a pair of dice the sum can not equal both 7 and 11 it can only equal one of
these at a time. For mutually exclusive events you can simply add the probabilities
together.
Example:
A) A hat contains 10 red marbles, 5 blue marbles, 3 pink marbles and 2 black
marbles, find the probability that you pick a blue OR pink marble.
You know the events are mutually exclusive because you can not pick a marble that is
both blue and pink so you add the probability of getting a blue marble to the probability
of getting a pink marble.
P(blue) = 5/20
P(pink) = 3/20
P(blue OR pink) = P(blue) + P(pink) = 5/20 + 3/20 = 8/20
(5) Find the probability of events that are not mutually exclusive.
To do this you may need to use the Additive Rule:
P( E U F ) = P( E ) + P( F ) − P( E I F )
Example: Given P(E) = .56 and P(F) = .64 and P( E I F ) = .35 find P ( E U F )
P( E U F ) = .56 + .64 -.35 = .85
You can also use Venn Diagrams to calculate probabilities (see Example 5 pg 380).
(6) Calculate the probability of the complement of an event
The complement of an event is when anything but that event occurs. The complement of
event E occurring is not E occurring, it is denoted as E .
P( E ) = 1 – P(E)
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(7) Probability Using Combinations (Section 7.3)
A) Defective problems or problems involving the probability of being dealt a
certain hand in cards.
Example: A box contains twelve batteries, 7 are defective. If you pick a random sample
of 5 batteries what is the probability that exactly 3 are defective?
P(3 are defective) =
C (7,3) × C (5, 2)
= .44
C (12,5)
From the 7 defective batteries you want to pick 3 and from the 5 good batteries you want
to pick 2.
B) You are dealt a hand of 13 cards, find the probability that you have 4
hearts, 3 spades, 5 diamonds, and 1 club.
To solve this problem you should know that there are 52 cards in a deck of
cards – each deck contains 13 hearts, 13 spades, 13 diamonds, and 13 clubs.
From the 13 hearts you want 4, from the 13 spades you want 3, from the 13
diamonds you want 5, and from the 13 clubs you want 1.
P (4hearts,3spades,5diamonds,1c lub) =
C (13, 4) × C (13,3) × C (13,5) × C (13,1)
= .00539
C (52,13)
(8) Conditional Probability (Section 7.4)
Let E and F be events of a sample space S and suppose P(F) >0 . The conditional
probability of event E, assuming event F has occurred, is denoted by P ( E F ) and is
defined by:
P( E F ) =
P( E ∩ F )
P( F )
The Product Rule – for two events E and F, the probability of event E and F occurring
( P ( E ∩ F ) ) is :
P( E ∩ F ) = P( F ) × P( E F )
You should know how to calculate conditional probability using the conditional
probability formula and tree diagrams.
Examples:
4
1. If P(F) = 2/3 and the P ( E ∩ F ) = ¼, what is the probability of event E occurring
given event F has occurred:
1
P( E ∩ F ) 4 3
P( E F ) =
= =
2 8
P( F )
3
2. Given the following tree diagram find the probabilities:
.45
.6
C
A
.55
.4
.3
B
.7
D
C
D
P(C)= .60*.45 + .40*.30
P (C A) = .45
P (C B ) = .3
(9) Independent Events (Section 7.5)
Two events (event E & F) are independent of each other if the occurrence or
nonoccurrence of one does not change the probability that the other will occur. This
means that,
P( E F ) = P( E )
P( F E ) = P( F )
If two events are independent then,
P ( E ∩ F ) =P(E) * P(F)
You can use these equations if you know events are independent and also to prove if
events are independent or not.
(10)
Binomial Probability (Section 8.2)
Any random experiment for which the binomial probability model is appropriate is called
a Bernoulli Trial. Random experiments are called Bernoulli trials if:
(a) The same experiment is repeated several times
(b) There are only two possible outcomes, success and failure,
on each trial.
(c) The repeated trials are independent
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(d) The probability of each outcome remains the same for each
trial.
To use the binomial probability model your experiment must have all four
characteristics of the Bernoulli Trial – if it does you can use the following equation
to calculate probability.
b(n, k ; p) = C ( n, k ) × p k q ( n − k ) =
n!
p k q( n−k )
k !(n − k )!
where,
n = total # of trials
k = # of successes
p = probability of success
q = probability of failure = 1-p
Typical problems where you can use the binomial probability model include flipping a
coin, rolling dice, true-false or multiple choice tests, and product testing.
Example: If a fair coin is flipped 3 times find the probability of getting exactly 2 heads
Use the binomial probability equation where n= 3, k = 2, n-k=1,p = 1/2 , and q = ½
P(exactly 2 heads) = C (3, 2) × .502 × .501 = .375
(11)
Expected Value (Section 8.3)
Expected Value = E = m1 × p1 + m2 p2 + .........mn × pn
Example:
Outcome
E1
E2
Probability
.20
.25
Payoff
3
2
E3
.30
1
E4
.25
-2
Calculate the expected value = E = 3(.20) + 2(.25) + 1(.30) + (-2)(.25) = .90
(12)
Random Variable (Section 8.5)
A random variable is a rule that assigns a number to each outcome of an experiment. For
instance, if a coin is flipped 2 times let X be a random variable equal to the number of
heads obtained then,
Sample Space
HH
HT
TH
TT
X
2
1
1
0
And the probability distribution is :
X
Probabiilty
0
.25
1
.50
2
.25
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Sample Problems:
(1.) Define the sample space for flipping a coin and then rolling a die. What is the
size of this sample space?
(2) Define the sample space for spinning a wheel with the numbers 1 through4 and
then spinning a wheel with the letters A,B,C on it. What is the size of this sample
space?
(3) Define the sample space for a three questions true –false test. What is the size of
this sample space?
(4) Define the sample space for tossing a coin 4 times by constructing a tree diagram.
(5) Is a probability of 1.2 possible?
(6) If an event has 3 probable outcomes are the following probabilities valid for this
event? P(1) = .42, P(2) = .22, and P(3) = .40
(7) A box contains 5 pink marbles, 3 red marbles, 2 black marbles, and 7 orange
marbles.
a. What is the probability that you pick a red marble?
b. What is the probability that you pick a red OR black marble?
c. What is the probability you pick a green marble?
(8) A pair of dice is rolled once. Find the following probabilities for the sum of the
dice:
a. P(7)
b. P(4)
c. P(7 OR 4)
d. P(sum is even)
e. P(Sum is even OR greater than 7)
f. Draw a Venn Diagram representing the probabilities in (e)
(9) Sheldon is taking courses in both Math and History. He estimates his probability
of passing Math at 62% and the probability of passing History at 45%. He also
estimates his probability of passing at least one course at 75%. Use a Venn
Diagram to figure out what the probability is that Sheldon passes both courses.
(10)
Let E and F be events of a sample space S and let P(E)=.45, P(F) = .35,
and P( E ∩ F ) = .3 . Find the probabilities of the following events:
7
a.
b.
c.
d.
E or F
F but not E
E but not F
Neither event
(11)
A single card is drawn at random form a deck of 52 cards. Find the
following probabilities:
a. P(a heart)
b. P(a 7 of diamonds)
c. P(a picture card)
d. P(black card)
e. P(black card or a picture card)
(12)
A box contains fifteen light bulbs of which 6 are defective. If you pick a
sample of three light bulbs find the probability that:
a. Exactly two are defective.
b. No light bulbs are defective.
c. At least two light bulbs are defective.
(13)
You roll a pair of dice 3 times. Using the binomial probability, what is the
probability you get (hint success = sum of 6, failure = sum not=6)
a. The sum of six all three times
b. The sum of six at least twice
c. The sum of 6 exactly once
(14)
Two cards are drawn from random from an ordinary deck of playing cards
find the probability that:
a. Both cards are aces.
b. Both cards are red.
c. At least one card is black.
(15)
Five cards are dealt at random from a regular deck of 52 playing cards.
Find the probability that
a. All are spades
b. Exactly 3 are red
c. Exactly 4 are clubs
P(E) = .4 P(F) = .52 P ( E ∪ F ) = .65 , find:
(16)
a. P ( E ∩ F ) – hint use the Additive Rule
b. P ( E F )
c. P ( E F )
d. P ( F E )
Two cards are drawn at random(without replacement) from a regular deck
(17)
of 52 cards. What is the probability that
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a. The first card is a spade and the second card is black(hint: you are looking
for (P(1st card is spade) ∩ P(2nd card is black))
b. The first cards is black and the second card is a spade (hint: you are
looking for (P(1st card is black) ∩ P(2nd card is spade))
(18)
From a box containing 4 red, 5 white, and 6 green balls, 2 balls are drawn
at a time without replacement. Find the probability that 1 red and 1 green ball are
drawn.
(19)
If E and F are independent events with P(E) = .42 and P(F) = .30 find,
a. P ( E F )
b. P ( F E )
c. P ( E ∩ F )
d. P ( E ∪ F )
(20)
Ian has just been given a 10-question multiple choice quiz in Science
class. Each question has five possible answers, of which only one is correct. Ian
didn’t study for his test and guesses at all the answers.
a. What is the probability he will answer all the questions correctly?
b. What is the probability he will answer all the questions incorrectly?
c. If Ian needs to get at least 7 correct to get at least a B, what is the
probability that Ian gets at least a B?
(21)
A contestant on the game show Deal or No Deal has 5 suitcases left to
pick from – each suitcase has some money in it – if the contestant picks one of the
remaining suitcases what is the expected value given:
Suitcase
Money ($)
1
.01
2
750,000
3
1,000
4
50,000
5
300,000
(22)
An urn contains 3 blue and 5 white balls. Two balls are drawn with
replacement. Let random variable X equal the number of blue balls drawn, list the
values of X and the probability distribution. (Hint: use binomial distribution)
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