Download Probability The Bag Model Imagine a bag (or box) containing balls

Probability
The Bag Model
Imagine a bag (or box) containing balls of various kinds–having
various colors for example. Assume that a certain fraction p of these
balls are of type A. This means
N = total number of balls
NA = total number of balls of type A
NA
and so p =
N
We draw a ball from the bag and say:
The probability of drawing a ball of type A is p.
If the bag contains red and blue balls, two-fifths of which are red,
then the probability of drawing a red ball is two out of five, or
p=
2
= .4 = 40%
5
1
2
First Principles
The number NA of balls of type A cannot be negative and cannot
exceed the total number of balls in the bag. That is,
0 ≤ NA ≤ N
Dividing by N we get
0
NA
N
≤
≤
N
N
N
or
0 ≤ Probability of type A ≤ 1
If NA counts the number of balls of type A, then N − NA counts
the number of balls which are not of type A. Thus the probability
that a ball chosen at random is not of type A is
N − NA
N
NA
=
−
= 1 − probability of type A
N
N
N
If there is a 40% chance of rain, then there is a 60% it won’t rain.
3
Reducing Problems to the Bag Model
Heads or Tails
Assuming the coin is “fair.” The bag contains one ball bearing
the inscription “heads” and one ball bearing the inscription “tails.”
The probability of heads is 21 and the probability of tails is 12 .
The die (plural of die is dice)
With a fair die the probability of throwing a 3 is 61 . What does
this mean in the bag model?
The bag contains 6 balls, bearing the inscriptions 1, 2, 3, 4, 5,
and 6. A throw of the die corresponds to drawing one of the six
balls from the bag. The probability of throwing a 3 is
p=
N3
1
=
N
6
Rain
A weather forecaster says “There is a 40% chance of rain.” This
means put 40 red balls in the bag (red for rain) and 60 blue balls
(blue for blue skies). A cloud choses a ball from the bag. A red ball
causes the cloud to release precipitation.
4
Birth Day
The probability of being born on a Monday is 1/7. The bag contains slips of paper bearing the names of the days of the week. The
slip of paper drawn from the bag by the baby-to-be-born determines
its birth day.
Gender Selection
The probability of the baby being a boy is 12 . (Actually it is
somewhat higher: 51.4%.) On fertilization, the ovum draws a slip of
paper from the bag containing the slips marked “boy” or “girl.” (In
effect, the sperm can be considered as bearing the two inscriptions;
male sperm being somewhat swifter account for the discrepancy
from 50%.)
Death
The probability of dying at some time or other is 100%. For a
Dutchman the probability of dying within one year is 0.77%. Such
is life: out of 10,000 balls, 77 are colored red and 9923 are blue.
When we meet someone in the steet, we draw a ball from the bag.
A red ball says “dead within a year,” a blue ball says “she will live
a year longer.”
5
Cards
There are 52 cards in a standard deck of cards (not counting the
jokers). There are four suits:
Diamond ♦
Heart ♥
Club ♣
Spade ♠
Each suit contains 13 cards:
Ace 2
4 · · · 9 10
3
Jack
Queen King
A card is drawn from a well-shuffled deck. The probability that the
card is
• the queen of hearts is
• a heart is
• a queen is
13
52
4
52
=
=
1
52
1
4
1
13
• a heart or a queen is
?
52
6
A pair of dice
36 throws are conceivable, namely the pairs (f irst, second) where
the first die shows f irst dots and the second die shows second dots.
These outcomes could be arranged in a 6 by 6 grid:
(1, 6)
(2, 6)
(3, 6) (4, 6)
(5, 6)
(6, 6)
(1, 5)
(2, 5)
(3, 5)
(5, 5)
(6, 5)
(1, 4)
(2, 4)
(3, 4) (4, 4) (5, 4)
(6, 4)
(1, 3)
(2, 3)
(3, 3) (4, 3) (5, 3)
(6, 3)
(1, 2)
(2, 2)
(3, 2)
(4, 2)
(5, 2)
(6, 2)
(1, 1)
(2, 1)
(3, 1)
(4, 1)
(5, 1)
(6, 1)
(4, 5)
7
Two dice are thrown. Find the probability that
• the first die is a 5
• the second die is a 5
• both dice are 5’s
• at least one die is a 5
• the two dice are the same (doubles)
• the sum is 2 (snake-eyes)
• the sum is 9
• the sum is 7 (craps)
8
Mutually Exclusive Events
Two events are called mutually exclusive if they cannot both
occur simultaneously.
Examples:
Catholic and Protestant
Male and Female
Heads and Tails
Rule for mutually exclusive events: When two events are mutually exclusive, then the probability that one or the other will occur
is the sum of the two probabilities.
Balls in a Bag. There are 5 green, 3 blue, and 2 red balls in a bag.
The probability that a ball chosen at randon is green or red is the
probability the ball is green + the probability it is red
=
5
2
7
+
=
10 10
10
Single throw of a die. A single die is thrown. The probability the
number on top is a 5 or a 6 is
=
1 1
2
1
+ = =
6 6
6
3
9
Two events are called non mutually exclusive if it is possible
for both to occur.
Examples:
Catholic and Male
Lawyer and Liar
Ace and Club
Rule for non mutually exclusive events: When two events are
mutually exclusive, then the probability that at least one of them
occurs is the sum of the two probabilities minus the probability that
they both occur.
Cards. A card is drawn from a standard deck. The probability it is
an ace or a club equals The probability it is an ace + the probability
it is club − the probability it is both
=
13
1
16
4
+
−
=
52 52 52
52
Two dice. A pair of dice is thrown. The probability that at least
one die is a 5 is
6
1
11
6
=
+
−
=
.
36 36 36
36
10
Independence
Two events are called independent if the occurrence or nonoccurrence of one event in no way affects the probability of the second
event.
Rule for independent events: When two events are independent,
then the probability that they both occur is the product of their
separate probabilities.
Flipping a coin two times. The outcome of the second toss is in no
way dependent on the outcome of the first toss. So the probability
of landing two heads in a row is the probability of heads on the first
toss times the probability of heads on the second toss
=
1 1
1
· = .
2 2
4
Another way to see this is to list the possible outcomes:
first toss
second toss
Heads
Heads
Heads
Tails
Tails
Heads
Tails
Tails
11
Two events are not independent if the probability of one event
depends on the occurrence or nonoccurrence of the other event.
Cards. Two cards are drawn from a standard deck without replacing the first card. The probability that the second card is an ace
depends on what the first card was. If the first card was also an
ace, then 3 aces are left in a deck of 51 remaining cards. If the first
card was not an ace, then 4 aces are left in a deck of 51 cards.
Rule for non independent events: When two events are not
independent, then the probability that they both occur is the product of the probability the first event occurs times the probability
the second event occurs assuming that the first has occurred.
Cards. Draw two cards from a deck (without replacing the first
card). The probability that both cards are aces is the probability
that the first card is an ace times the probability the second card is
an ace assuming that the first was an ace
=
12
4·3
1
4 3
·
=
=
=
.
52 51
52 · 51
4 · 13 · 3 · 17
13 · 17
12
Probability and Normal Distribution
Suppose a data set is normally distriubted with a mean of
µ = 140
and a standard deviation of
σ = 6.
What is the probability that an element chosen at random from the
data set will lie in the range
128
to
155?
Solution: We need to figure the z-values.
How many standard deviations to the left of µ is 128?
µ − zσ = 128
140 − z · 6 = 128
140 − 128 = 6z
6z = 12
12
z=
=2
6
13
How many standard deviations to the right of µ is 155?
µ + zσ = 155
140 + z · 6 = 155
6z = 155 − 140
6z = 15
15
= 2.5
z=
6
So the range 128 to 155 represents 2 standard deviations below the
mean to 2.5 standard deviations above it.
The z-table gives the percentages 47.7% and 49.4% corresponding
to z = 2 and z = 2.5. So the probability the data element lies
between 128 and 155 is
47.7 + 49.4 = 97.1%