Chapter 15. Probability, continued
Review:
The sample space of a random experiment is the set of all possible
outcomes of the experiment.
We have to count outcomes; a useful tool is the multiplication
rule: count outcomes in two or more stages; suppose in each
stage the number of choices doesn't depend on choices made in
other stages; then the number of outcomes is the product of the
number of choices in each stage.
Example: five candidates in an election, the one with the most
votes is President, 2nd most is VP, 3rd most is Secretary. Outcomes:
Section 3. Permutations and combinations
A standard deck of cards has four suits (Hearts, Diamonds, Spades
and Clubs) and 13 values (Ace, King, Queen, Jack, 10, 9, 8, 7, 6, 5,
4, 3, 2), making 4 * 13 = 52 cards. A poker hand consists of five
cards. How many different poker hands are there?
There are 52 * 51 * 50 * 49 * 48 = 311,875,200 hands, if we pay
attention to the order in which the cards are dealt.
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In general, if you have n items, and you choose r of them, and you
care about the order in which you choose them,
then the number of choices is called the
number of permutations on n objects taken r at a time =
nPr = n * (n – 1) * (n – 2) * ... * (n – r + 1)
(notice that there are r numbers being multiplied).
So the number of 4 member committees (in which order matters)
from 35 people = 35P4 = 35 * 34 * 33 * 32.
The number of 5 card hands (in which order matters) from 52 cards
= 52P5 = 52 * 51 * 50 * 49 * 48.
In draw poker, the order we receive the cards doesn’t matter.
When we counted the ordered hands above, we counted each
unordered hand many times.
How many ordered hands does each unordered hand give rise to?
There are 5 * 4 * 3 * 2 * 1 = 5! = 120 orderings of a five card
hand. So the number of unordered hands is
52 * 51 * 50 * 49 * 48)/( 5 * 4 * 3 * 2 * 1) =
311,875,200/120 = 2,598,960
In general, if you have n items, and you choose r of them, and you
don’t care about the order in which you choose them, then the
number of choices is called the
number of combinations of n objects taken r at a time =
n
Cr = nPr / r! =
[n * (n – 1) * (n – 2) * ... * (n – r + 1)]/[r * (r – 1) * (r – 2) *...*1]
2
Suppose there are 10 horses entered in a race.
In how many ways can one pick
a) the top three finishers regardless of order?
b) the first-, second-, and third-place finishers in the race?
b) asks for the number of permutations = 10P3 = 10 * 9 * 8 = 720.
a) asks for the number of combinations = 10C3 = 720/3! = 720/6 =
120.
Look at book's problems 33-36 and determine if you are asked for
permutations or combinations
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Sections 4. Probability spaces
The probability of something happening is a number between 0
and 1; if a random experiment is carried out repeatedly, the
probability of an outcome is supposed to represent the fraction of
the experiments for which that outcome is expected to occur. The
book gives the example of a basketball player who, over a long
period of time, shoots 2135/2362 = .904 of his free throws. So one
might conclude that if he has a free throw attempt (that’s the
random experiment in this case), .904 is the probability of success
and 1 - .904 = .096 is the probability of failure.
In general, for some random experiment, you must first determine
the sample space S = {o1, ... ,oN}. Next you must assign a
probability Pr (oi) to each outcome oi . Each probability is between
0 and 1, and Pr (o1) + … + Pr (oN) = 1. This is called a probability
assignment.
If we toss a fair coin two times, and note on each toss whether it is
heads or tails, then the sample space S = {HH, HT, TH, TT}.
Since each of these outcomes is equally likely,
Pr (HH) = Pr (HT) = Pr (TH) = Pr (TT) = ¼ = .25 .
If we toss a fair coin two times, and note the number of heads, then
the sample space S = {2, 1, 0}. But these outcomes are not equally
likely, since HT and TH both result in an outcome of 1. So
Pr (2) = Pr (0) = .25, and Pr (1) = .5 .
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p580 #38b. Consider the sample space S = {o1, o2, o3, o4}.
Suppose you are given Pr (o1) + Pr (o2) = Pr (o3) + Pr (o4).
If Pr (o1) = .15 and Pr (o3) = .22, find the probability assignment.
(So we need to find Pr (o2) and Pr (o4)).
Answer: Pr (o1) + Pr (o2) = Pr (o3) + Pr (o4) = .5
(since they are equal and add to 1), so
Pr (o2) = .5 - .15 = .35 and Pr (o4) = .5 - .22 = .28.
There are 8 players (call them P1, … , P8) entered in a chess
tournament. Our random experiment is to hold the tournament and
note the winner. The sample space is S = { P1, … , P8}. We are
told that P1 has a 25% chance of winning, P2 has a 15% chance, P3
has a 5% chance, and all others have an equal chance. Find the
probability assignment.
Answer: We are given that
Pr (P1) = .25, Pr (P2) = .15 and Pr (P3) = .05.
So the probability of one of these three players winning is
Pr (P1) + Pr (P2) + Pr (P3) = .25 + .15 + .05 = .45.
Thus the probability of one of the other players winning is
Pr (P4) + Pr (P5) + Pr (P6) + Pr (P7) + Pr (P8) = 1 - .45 = .55 .
Also, Pr (P4) = … = Pr (P8), so each is 1/5 of .55, that is
Pr (P4) = … = Pr (P8) = .55/5 = .11 .
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Events.
An event is any subset of the sample space. The probability of an
event is the sum of the probabilities of the outcomes in that event.
If we toss a fair coin two times, and note on each toss whether it is
heads or tails, then the sample space S = {HH, HT, TH, TT}. Here
is a listing of some of the events:
Description of event
Toss two heads
Toss one head
Toss at least one head
Toss two heads or two tails
Toss at most two heads
Toss three heads
Set of outcomes
{HH}
{HT, TH}
{HH, HT, TH}
{HH, TT}
{HH, HT, TH, TT}
{}
Probability
.25
.5
.75
.5
1
0
The last two events listed are called the certain event and the
impossible event.
p.580 # 44. Consider the random experiment where a student takes
a four-question true-false quiz. Write out the event described by
each of the following statements:
a) Exactly two of the answers given are T’s (T for true).
b) At least two of the answers given are T’s.
c) At most two of the answers given are T’s.
d) The first two answers given are T’s.
Answers: a) {TTFF, TFTF, TFFT, FTTF, FTFT, FFTT}
b) {TTFF, TFTF, TFFT, FTTF, FTFT, FFTT, TTTF, TTFT, TFTT,
FTTT, TTTT}
c) {FFFF, TFFF, FTFF, FFTF, FFFT, TTFF, TFTF, TFFT, FTTF,
FTFT, FFTT}
d) {TTFF, TTTF, TTFT, TTTT}
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