Transcript Introduction - Washington University in St. Louis

Chance
We will base on the frequency theory to study
chances (or probability).
Definition
• The chance of something gives the
percentage of time it is expected to
happen, when the basic process is done
over and over again, independently and
under the same conditions.
Examples
• One simple game of chance involves
betting on the toss of a coin.
• The process of tossing the coin can be
repeated over and over again,
independently and under the same
conditions.
• The chance of getting heads is 50%: in the
long run, heads will turn up about 50% of
the time.
Examples
• A die is a cube with 6 faces, labelled: 1, 2, 3, 4,
5, 6.
• When we roll a die, the faces are equally likely to
turn up.
• The chance of getting a 1 is 1 in 6, i.e. 16.7%.
• This means that if the die is rolled over and over
again, repeating the basic chance process under
the same conditions, in the long run a 1 will
show about 16.7% of the time.
Properties of chances
• If something is impossible, it happens 0% of the
time.
• If something is sure to happen, then it happens
100% of the time.
• Chances are between 0% and 100%.
• The chance of something equals 100% minus
the chance of the opposite thing.
• For example, suppose we are playing a game,
and have a 45% chance to win. This means we
expect to win about 45% of the time, and we
also must expect to lose the other 55% of the
time.
Compute the chances
• Based on the frequency theory, the
chance of some event is the ratio:
• The frequency of some event / the total
number of the events.
Example
• A box contains red marbles and blue marbles.
One marble is drawn at random from the box
(each marble has an equal chance to be drawn).
If it is red, you win $1. If it is blue, you win
nothing. You can choose between two boxes:
• Box A contains 3 red marbles and 2 blue ones;
• Box B contains 30 red marbles and 20 blue
ones.
• Which box offers a better chance of winning, or
are they the same?
Solution
• The two boxes offer the same chance of
winning.
• Based on the frequency theory, in the long
run, in box A, each of the 5 marbles will
appear about 1 time in 5. So the red
marbles will turn up about 3/5 of the time.
The “3” is the “frequency” of red marbles,
and the “5” is the “total” number of the
events. So the chance is 60%.
Solution
• Similarly, in box B, each of the 50 marbles
will turn up about 1 time in 50. So the red
marbles will turn up about 30/50 of the
time, your chance of winning is again 60%.
• So by comparing the ratios, we know that
the chance of winning is the same.
Two main models
• 1. Draw two tickets at random with replacement
from the box: 1, 2, 3, … (Suppose in the box, we
mark the tickets as 1, 2, 3, …)
• This asks you to do the following process: shake
the box, draw out one ticket at random (equal
chance for all tickets), mark down the number,
put it back in the box, shake again, draw a
second ticket at random, mark down the number,
and put it back in the box.
Two main models
• 2. Draw two tickets at random without
replacement from the box: 1, 2, 3, …
• This asks you to do the following process: shake
the box, draw out one ticket at random, set it
aside, draw out a second ticket at random.
Remark
• When you draw at random, all the tickets
in the box have the same chance to be
picked.
• The two models can be applied to more
general cases: e.g. (1) draw tickets at
random from the box with more tickets---30, 50, 100 or more tickets; (2) draw more
tickets at random----draw 5, 10, 100 or
more tickets from the box.
The multiplication rule
The chance that two things will both happen
equals the chance that the first will happen,
multiplied by the chance that the second will
happen given the first has happened.
Example
• A box has three tickets, colored red, white
and blue: R, W, B.
• Two tickets will be drawn at random
without replacement. What is the chance
of drawing the red ticket and then the
white?
Solution
• Since we draw tickets at random, each of the
ticket has the same chance to be picked. So on
the first draw, the chance of drawing R and left
with W and B in the box is 1/3.
• On the second draw, the chance of drawing W is
½, since there are only 2 tickets remaining in the
box.
• So the chance of drawing R and then W is 1/3 x
½ = 1/6, or 16.7%.
Remarks
• If we exhaust all the possibilities of
drawing 2 tickets from the box:
• R & W, R & B, W & B, W & R, B & R, B &
W.
• We see that the chance is 1/6. This
coincides the result that we got.
Remarks
• We have to be cautious to use the arguments in the
above solution.
• Because the chance or probability means the
percentage of time it is expected to happen. It
represents the percentage in the long run, not just
one time.
• A better understanding could be: suppose you start
with 600 people, each of these people holds a box:
R, W, B. About 200 of them will get R on the first
draw. Of these 200 people, about 100 will get W on
the second. So the chance is 100/600 = 1/6 =
16.7%.
More examples
• Two cards will be dealt off the top of a wellshuffled deck. What is the chance that the first
card will be the seven of clubs and the second
card will be the queen of hearts?
• Note. A deck of cards has 4 suits: clubs,
diamonds, hearts, spades. There are 13 cards in
each suit: 2 through 10, jack, queen, king, ace.
So there are 4 x 13 = 52 cards in total.
More examples
• Solution.
• The chance that the first card will be the seven
of clubs is 1/52.
• Given the first card was the seven of clubs, there
is 51 cards remaining, so the chance that the
second card will be the queen of hearts is 1/51.
• Therefore, the chance of getting both cards is
1/52 x 1/51 = 1/2652 ≈ 0.04%.
More examples
• A deck of cards is shuffled, and two cards are
dealt. What is the chance that both are aces?
• Solution.
• The chance that the first card is ace equals 4/52.
• Given the first card is an ace, there are 3 aces in
the 51 remaining cards. So the chance of a
second ace equals 3/51.
• So the chance that both are aces is 4/52 x 3/51
= 12/2652 ≈ 0.45%.
More examples
• A coin is tossed twice. What is the chance of a
head followed by a tail?
• Solution.
• The chance of a head on the first toss equals ½.
• No matter how the first toss turns out, the
chance of tails on the second toss equals ½.
• So the chance of heads followed by a tail is ½ x
½ = ¼.
Conditional probabilities
The chance of the event given something has
already happened.
Example
• A deck of cards is shuffled and the top two
cards are put on a table, face down. You
win $1 if the second card is the queen of
hearts.
• (a) What is your chance of winning the
dollar?
• (b) You turn over the first card. It is the
seven of clubs. Now what is your chance
of winning?
Solution
• (a) Note that the bet can be settled without
even looking at the first card. The second
card is all you need to know.
• When the deck is shuffled, this brings the
cards into random order. The queen of
hearts has to be somewhere. There are 52
possible positions. So there is 1 chance in
52 for this card to be the second. Hence
the chance of winning is 1/52.
Solution
• (b) Now the first card is already turned
over. This has already happened.
• There are 51 cards left, and they are in
random order.
• So event that the queen of hearts is the
second one has the chance 1 in 51.
• Hence the chance of winning is 1/51.
Remark
• In part (b), 1/51 is a conditional probability (or
conditional chance). The problem puts a
condition on the first card.
• The conditional probability reads the second
card is the queen of hearts given the first card is
the seven of clubs. (Format: event A given event
B, as we already seen on previous examples.)
• In part (a), 1/52 is an unconditional probability
(or unconditional chance). There is no condition
on it.
Formula for conditional
probability
• Suppose we have two events: event A and
event B.
• We want to compute the conditional
probability for event A given event B.
• Then the formula for this will be:
• P(A|B) = P(A, B) / P(B).
• In fact, this is a reformulation of the
multiplication rule:
• P(A, B) = P(A|B) x P(B).
Formula for conditional
probability
• For example, in part (b) of our previous problem, the
conditional probability can be expressed as:
• P(2nd card is queen of hearts | 1st card is seven of clubs).
• In part (a), the unconditional probability reads:
• P(2nd card is queen of hearts).
• Note: Sometimes, it is easy to figure out P(A|B) as part
(b) in the problem, so we may apply the multiplication
rule to calculate P(A, B). (See the previous examples.)
• Sometimes, it will be easy to find P(A, B), then we could
apply the formula for conditional probability to compute
P(A|B).
Formula for conditional
probability
• By using the formula, we may obtain the result in
part (a) in an alternative way:
• P(2nd card is queen of hearts) = P(2nd card is
queen of hearts | 1st card is others) x P(1st card
is others) = 1/51 x 51/52 = 1/52.
• There are 51 chances in 52 when 1st card is
others. This is where 51/52 comes from.
• Given 1st card is others, the conditional
probability is 1/51. Hence we obtain the result.
Independence
Definition
• Two events are independent if the
chances (or probabilities) for the second
given the first are the same, no matter how
the first one turns out.
• Using an equation, we can express:
• P(A|B) = P(A), or equivalently:
• P(A, B) = P(A) x P(B).
• Otherwise, the two events are dependent.
Example
• Someone is going to toss a coin twice. If
the coin lands heads on the second toss,
you win a dollar.
• (a) If the first toss is heads, what is your
chance of winning the dollar?
• (b) If the first toss is tails, what is your
chance of winning the dollar?
• (c) Are the tosses independent?
Solution
• (a) If the first toss is heads, there is a 50%
chance to get heads the second time.
• (b) If the first toss is tails, the chance is still
50%.
• (c) No matter how the first toss turns out,
the chances for the second toss stay the
same: 50%. So they are independent.
Example
• Two draws will be made at random with
replacement from a box of tickets:
1,1,2,2,3 (tickets marked by numbers).
• (a) Suppose the first draw is 1. What is the
chance of getting a 2 on the second draw?
• (b) Suppose the first draw is 2. What is the
chance of getting a 2 on the second draw?
• (c) Are the draws independent?
Solution
• Since we draw tickets with replacement,
the second draw is always made from the
same box with 1,1,2,2,3.
• So no matter what we draw on the first,
the chance of getting 2 on the second
stays the same: two in five, i.e. 2/5 = 40%.
• This shows the draws are independent.
Example
• If we modify the previous example a little
bit: the draws are made without
replacement. Are the answers stay the
same?
Solution
• (a) If the first draw is 1, then before the
second draw, the box has tickets: 1,2,2,3.
So the chance for drawing 2 becomes
50%.
• (b) If the first draw is 2, then before the
second draw, the box has tickets: 1,1,2,3.
So the chance for drawing 2 becomes
25%.
• (c) By above arguments, the draws are
dependent.
Remark
• When drawing at random with replacement, the draws
are independent.
• Without replacement, the draws are dependent. (Exclude
some extreme cases.)
• When we have a large number of tickets (say 10,000),
we can treat the draws are independent even though we
draw tickets without replacement. (The probabilities only
change a little.)
• When the numbers on the tickets are identical (say
identically 1), then draws are independent even though
we draw tickets without replacement.
Apply the formula
• Example:
• A box has three tickets, colored red, white,
and blue. (Mark as R,W,B.)
• Two tickets will be drawn at random with
replacement. What is the chance of
drawing the red ticket and then the white?
Apply the formula
• Solution.
• Recall that if two events are independent,
then:
• P(A, B) = P(A) x P(B).
• Since we draw tickets with replacement,
they are independent. So the probability is
• 1/3 x 1/3 = 1/9.
Remark
• If we compare this example to the
example on slide 14, we find that it is
easier this time. We don’t need to work
with the conditional probabilities.
• This is how independence matters.
• The independence formula is a special
case of the multiplication rule.
Summary
• The probability / chance of something
gives the percentage of times the thing is
expected to happen, when the basic
process is repeated over and over again.
• Probabilities are between 0% and 100%.
Impossibility is represented by 0%,
certainty by 100%.
• The chance of something equals 100%
minus the chance of the opposite thing.
Summary
• The multiplication rule: P(A, B) = P(A|B) x
P(B).
• Two events are independent if the
chances for the second one stay the same
no matter how the first one turns out:
P(A|B) = P(A).
• Consequence of independence: P(A, B) =
P(A) x P(B).
Summary
• When we draw tickets at random, all
tickets in the box share the same chance
to be picked.
• Draws made at random with replacement
are independent.
• Without replacement, the draws are
dependent.