### discrete

```AP Statistics B
February 29, 2012
Leap year day! Not another
for another 4 years!
AP Statistics B warm-up
Wednesday, February 29, 2012
1. Do Problem 7, Chapter 16, p.382
2. What does the word “independently”
mean in the last sentence of
problem 7.
When you’re ready, go on to the next
slide.
Wednesday, February 29, 2012
1. What are the probabilities of getting the
contracts? 30% for the first, 60% for the
second. So the expected value is the product
of the probabilities times each contract value
(.3)(\$50,000)+(.6)(\$20,000)
=\$15,000+\$12,000=\$27,000.
2. “Independently” here means that the
probability of getting the second contract is
unrelated to the probability of getting the first.
Alternatively stated, winning the first contract
neither increases nor decreases the likelihood
that the company will win the second, smaller
contract.
Outline for today’s class
• Review problems 17-20, Chapter 15.
• Examine expected values
• Discrete v. continuous random
variables
lecture
• Don’t worry about writing everything
garfieldhs.org, look under “Statistics(all
presentation for today (colleges are
doing this a lot).
• Follow along the general path of the
ideas. You can sweat the details later.
• Give me feedback on what would make
the presentation better for you.
Before we begin, remember
the general multiplication rule
P( A  B)  P( A)  P( B | A)
Meaning of P(A)
• In problems involving choosing without
replacements, it’s the first draw
• Examples:
deck of 52, so 13/52 = ¼
– Odds of drawing an ace—4 aces out of 52,
so 4/52 = 1/13.
– Odds of drawing anything BUT a heart—13
hearts, so 52-13=39 other cards out of 52,
so probability is 39/52=3/4=0.75
Understanding P(B|A)
• (B|A) means that something has happened
to change the probability of B from
happening.
• In these cases, that “something” is that a
card has been drawn. It can have two
effects:
– It ALWAYS decreases the deck by 1 card, so
your denominator is now 51, not 52
– It MAY affect the remaining odds. For example,
if the first odds are drawing an ace, the aces in
the deck are reduced, and the total cards are
reduced by one, so the odds drawing the next
ace on the second hand is 3/51.
Subsequent probabilities may
be affected
• You could continue to draw cards, and
the probabilities will change and may
be affected by P(A) and subsequent
odds.
• Examples:
– Probability that the third card is NOT an
ace after drawing 2 aces
– Probability that you will draw a third space
in a row
• We will now examine how these work
with some of the homework problems.
Chapter 15, problem 17 (b-d)
(drawing three cards)
• Sometimes it’s better to verbalize
rather than look for formulas.
• With (b), all 3 cards you draw are
red.
• # of red cards in the deck: 26 (13
hearts, 13 diamonds)
• So for the first card you draw, what
are the odds that it will be red?
Problem 17(b)(drawing red
cards only)
• Probability = # of red cards available
divided by # of ALL cards available
• # of all cards= 52 on first draw (13
of each suit; no jokers [Jason
Romero doesn’t count as a joker])
• Probability of drawing a red card on
FIRST draw is therefore 26/52, or ½.
Drawing red cards
• How have things changed for the second
draw?
• How many red cards do we have left?
– We already drew one red card
– Therefore, we have 26-1=25 red cards still
available to be drawn.
• How many cards altogether do we have
left?
– Again, we drew one card, so we have 52-1=51
cards remaining
• Probability of drawing a red card on second
draw is 25/51
Drawing red cards (still!)
• Third draw is the same:
– We are down to 24 red cards
– There are a total of 50 cards left in the
deck (we already drew two red cards
• Probability of red card on third draw is
therefore 24/50=12/25, if you want to
simplify.
• Total probability is PRODUCT of the
three: (1/2)(25/51)(12/50 )=.118 =
11.8%
• OK, so you’ve now seen how to
approach this.
• Work in groups on 17(c): “You are
dealt a hand of three cards, one at a
time. Find the probability that….(c)
• After 8-10 minutes, report out, and
advance to the next slide for my
explanation.
Problem 17(c):
• Similar in approach to the last
problem.
• First draw: the probability of getting
no spaces is 39/52. (if you like, you
can look at this as the complement
of getting spades….you have a 13/52
chance of getting spades, and the
complement is therefore 1-13/52 =
1-¼ = ¾ = 0.75
Problem 17(c):
you get no spades (slide 2)
• For second draw, there are still 13
total (13+38), so probability = 38/51.
• Third draw: similar to second draw:
– 50 cards left
– Probability of no spades on third=37/50
• Probability of no spades =
(3/4)(38/51)(37/50)= 0.414 = 41.4%
17(d): draw at least one ace
• More complicated, because there are
many possible outcomes.
• You could draw 1 ace, 2 aces, or
three aces.
• The total probability will be the sums
of all the individual probabilities.
• Let’s look at the odds of drawing
exactly 1 ace first.
Drawing exactly 1 acs
Here’s a table that summarizes the
three possibilities of drawing exactly
one ace:
1
2
3
1st
draw
Ace
no ace
no ace
2nd
draw
no ace
Ace
no ace
3rd
draw
no ace
no ace
Ace
Analysis of probability for
drawing exactly one ace
• 1st possibility: A, followed by 2 non-aces:
(4/52)(48/51)(47/50)=well, we’ll see in
coming slides.
• 2nd possibility: you draw an ace only on the
2nd draw, with non-aces on draws 1 and 3
• 3rd possibility: you don’t draw an ace until
the 3rd draw, with non-aces in draws 1 and
2.
• Next, I’ll show you how to catalog these
possibilities and figure out the total odds.
Summary for all 7 possibilities
Pattern
A00
0A0
00A
0AA
A0A
AA0
AAA
1st draw
2nd draw 3rd draw
04/52
48/52
48/52
48/52
04/52
04/52
04/52
48/51
04/51
47/51
04/51
48/51
03/51
03/51
47/50
47/50
04/50
03/50
03/50
48/50
02/50
Summary of probabilities
Pattern
A00
0A0
00A
0AA
A0A
AA0
AAA
1st draw
0.076923
0.923077
0.923077
0.923077
0.076923
0.076923
0.076923
2nd draw 3rd draw
0.941176
0.078431
0.921569
0.078431
0.941176
0.058824
0.058824
0.94
0.94
0.08
0.06
0.06
0.96
0.04
Pattern
A00
0A0
00A
0AA
A0A
AA0
AAA
Total
1st draw
0.076923
0.923077
0.923077
0.923077
0.076923
0.076923
0.076923
2nd draw
0.941176
0.078431
0.921569
0.078431
0.941176
0.058824
0.058824
3rd
Product
draw
0.94
0.94
0.08
0.06
0.06
0.96
0.04
0.068054
0.068054
0.068054
0.004344
0.004344
0.004344
0.000181
0.217376
Problem 18 (similar to 17, often with
the complements, or flip sides)
• For example, 18(a) has “no aces.”
Flip side of 17(c), which was “no
• Draw 1: 48/52.
• Draw 2: 47/51
• Draw 3: 46/50.
• Multiply them all together and you
get………………….
Problem 18(b):
all heart(s)
• 13 hearts. Just keep taking them
away as you reduce the deck.
• Work on this at your table for a few
minutes and see what you come up
with. My analysis is on the next slide.
•
•
•
•
First odds of drawing hearts: 13/52.
Second draw odds: 12/51
Third draw odds: 11/50.
Altogether: (13/52)(12/51)(11/50) =
Problem 19
• 12 batteries altogether
– Therefore, 7 are good.
(a) 1st two you pick are good:
(7/12)(6/11)=0.318
(b) At least one of the first three works. Set
this up like you did the “at least one ace” of
17(d), except you’ll have different odds
(c) The first 4 all work:
(7/12)(6/11)(5/10)(4/9)= something small
(7.1% per the book)
What looks hard isn’t
• 19(d) sounds awful: “you have to
pick 5 batteries to find one that
works.”
• Sounds like something worse than
the “at least one ace” from problem
17.
• What it ACTUALLY means is “you
have to pick 4 batteries BEFORE you
find one that works.”
How to model 19(d)
• If you look at a representation of
G=good, it looks like this: BBBBG
• Easy, right?
• (5/12)(4/11)(3/10)(2/9)(7/8)
• The book says that this comes out to
0.009.
Problem 20
• Much like problems 17 and 18, except
now considerably more complex
• We have 20 shirts (but the problem
doesn’t tell you that directly) broken
down as follows:
– 4 medium
– 10 large
– 6 XL
• Before going on to the next slide,
calculate the initial odds of getting a
medium, or a large, or an XL.
•
•
•
•
Medium = 4/20 = 0.20
Large = 10/20 = 0.50
XL = 6/20 = 0.30
All add up to 1, right?
• Note that the problem requires that you
get 2 mediums, one for you and one for
though it’s not really ambiguous.)
Analysis of problem 20
• (a) says “first two are wrong sizes”, i.e.,
either L or XL. Together those are 16
(10Ls + 6XLs), so probability is
(16/20)(15/19) = 0.632
• (b) says “the first medium shirt you
grab is the third”. In other words , you
got the wrong sizes the first 2 times, so
you can use the results from (a). The
odds of getting a medium on the 3rd
grab is 4/18, so answer is
0.632(4/18)=0.140
20(c) and (d)
• “The first four shirts you grab are all
extra large”
• This one is easy:
(6/20)(5/19)(4/18)(3/17)= 0.003
• (d), by comparison, is quite
complicated: 17 outcomes by my
calculations
• I’ll give extra credit worth a homework
assignment for anybody who completes
it by next Monday
Chapter 16: Expected values
• Covered this in previous lecture.
• Now to get the fancy formula:
E(X)=∑x∙P(x)
• A more familiar form of the equation
would be to use subscripts as
follows: E(X)=∑xi∙P(xi)
Expected values from the
lottery of yesterday
• Let’s use this formula with the lottery
we did yesterday to see how it
works.
• E(X)=∑xi∙P(xi) =
5(.7)+10(0.1)+20(.2)= \$8.50
• Here, i=1 corresponds to the 14 \$5bills, i=2 to the 2 \$10-bills, and i=3
to the 4 \$20-bills.
New concept:
discrete v. continuous
• “Discrete” means that you can list all
the outcomes, e.g.
– SAT scores (all SAT scores are integers, so
you can list each of them)
– Heights of a population (difficult, but you
COULD list them
• “Continuous” means the variable can
take on an infinite number of outcomes