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Probability 2.0
Independent Events
• Events can be "Independent", meaning each
event is not affected by any other events.
• Example: Tossing a coin.
• Each toss of a coin is a perfect isolated thing.
• What it did in the past will not affect the current
toss.
• The chance is simply 1-in-2, or 50%, just like ANY
toss of the coin.
• So each toss is an Independent Event.
Dependent Events
• But events can also be "dependent" ... which
means they can be affected by previous events
...
• Example: Marbles in a Bag
• 2 blue and 3 red marbles are in a bag.
• What are the chances of getting a blue marble?
• The chance is 2 in 5
• But after taking one out the chances change!
Marbles in a Bag….
• But after taking one out the chances change!
• So the next time:
• if we got a red marble before, then the chance
of a blue marble next is 2 in 4
• if we got a blue marble before, then the
chance of a blue marble next is 1 in 4
Replacement
• Note: if we replace the marbles in the bag
each time, then the chances do not change
and the events are independent:
• With Replacement: the events are
Independent (the chances don't change)
• Without Replacement: the events are
Dependent (the chances change)
Exercise
• There is a 2/5 chance of pulling out a Blue
marble, and a 3/5 chance for Red
• What are the chances of drawing 2 blue
marbles?
Solution
•
The chances of drawing 2 blue marbles is 1/10
Conditional event
• The probability that event B occurs, given that
event A has occurred, is called a conditional
probability.
• The conditional probability of B, given A, is
denoted by the symbol P(B|A).
• In other words, event A has already
happened, now what is the chance of event
B?
• And in our case: P(B|A) = 1/4
• So the probability of getting 2 blue marbles is:
• And we write it as
• "Probability of event A and event B equals
the probability of event A times the
probability of event B given event A"
Exercise
• A bag contains 3 red marbles and 4 blue
marbles. Two marbles are drawn at random
without replacement.
If the first marble drawn is red, what is the
probability the second marble is blue?
• A box contains 5 green pencils and 7 yellow
pencils. Two pencils are chosen at random
from the box without replacement.
What is the probability they are both yellow?
Finding hidden data
• Using Algebra we can also "change the
subject" of the formula, like this:
• Start with: P(A and B) = P(A) x P(B|A)
• Swap sides: P(A) x P(B|A) = P(A and B)
• Divide by P(A): P(B|A) = P(A and B) / P(A)
"The probability of event B given event A
equals the probability of event A and event
B divided by the probability of event A
Example
• 45% of the children in a school have a dog,
30% have a cat, and 18% have a dog and a cat.
What percent of those who have a cat also
have a dog?
Bayes’ Theorem
• Bayes' theorem. Let A1, A2, ... , An be a set of
mutually exclusive events that together form
the sample space S. Let B be any event from
the same sample space, such that P(B) > 0.
Then,
P( Ak | B ) = P( Ak ∩ B )
• P( A1 ∩ B ) + P( A2 ∩ B ) + . . . + P( An ∩ B )
http://stattrek.com/probability/bayestheorem.aspx
• Note: Invoking the fact that P( Ak ∩ B ) = P( Ak
)P( B | Ak ), Baye's theorem can also be
expressed as
P( Ak | B ) = P( Ak ) P( B | Ak )
• P( A1 ) P( B | A1 ) + P( A2 ) P( B | A2 ) + . . . + P(
An ) P( B | An )
When to Apply Bayes’ Theorem
• Part of the challenge in applying Bayes' theorem involves
recognizing the types of problems that warrant its use. You should
consider Bayes' theorem when the following conditions exist.
• The sample space is partitioned into a set of mutually exclusive
events { A1, A2, . . . , An }.
• Within the sample space, there exists an event B, for which P(B) > 0.
• The analytical goal is to compute a conditional probability of the
form: P( Ak | B ).
• You know at least one of the two sets of probabilities described
below.
– P( Ak ∩ B ) for each Ak
– P( Ak ) and P( B | Ak ) for each Ak
Example1
• Marie is getting married tomorrow, at an
outdoor ceremony in the desert. In recent
years, it has rained only 5 days each year.
Unfortunately, the weatherman has predicted
rain for tomorrow. When it actually rains, the
weatherman correctly forecasts rain 90% of
the time. When it doesn't rain, he incorrectly
forecasts rain 10% of the time. What is the
probability that it will rain on the day of
Marie's wedding?
Solution
• The sample space is defined by two mutuallyexclusive events - it rains or it does not rain.
Additionally, a third event occurs when the
weatherman predicts rain. Notation for these
events appears below.
• Event A1. It rains on Marie's wedding.
• Event A2. It does not rain on Marie's wedding.
• Event B. The weatherman predicts rain.
Solution
• In terms of probabilities, we know the
following: P( A1 ) = 5/365 =0.0136985 [It rains
5 days out of the year.]
• P( A2 ) = 360/365 = 0.9863014 [It does not rain
360 days out of the year.]
• P( B | A1 ) = 0.9 [When it rains, the
weatherman predicts rain 90% of the time.]
• P( B | A2 ) = 0.1 [When it does not rain, the
weatherman predicts rain 10% of the time.]
Solution
• We want to know P( A1 | B ), the probability it will
rain on the day of Marie's wedding, given a
forecast for rain by the weatherman. The answer
can be determined from Bayes' theorem, as
shown below.
• P( A1|B ) = _____P ( A1 ) P ( B | A1 )________
•
P( A1 ) P( B | A1 ) + P( A2 ) P( B | A2 )
• P( A1 | B ) = (0.014)(0.9) / [ (0.014)(0.9) +
(0.986)(0.1) ]
• P( A1 | B ) = 0.111
• The entire output of a factory is produced on
three machines. The three machines account
for 20%, 30%, and 50% of the output,
respectively. The fraction of defective items
produced is this: for the first machine, 5%; for
the second machine, 3%; for the third
machine, 1%. If an item is chosen at random
from the total output and is found to be
defective, what is the probability that it was
produced by the third machine?
• A solution is as follows. Let Ai denote the event that a
randomly chosen item was made by the ith machine (for i =
1,2,3). Let B denote the event that a randomly chosen item
is defective. Then, we are given the following information:
• P(A1) = 0.2, P(A2) = 0.3, P(A3) = 0.5.If the item was made
by machine A1, then the probability that it is defective is
0.05; that is, P(B|A1) = 0.05. Overall, we have
• P(B|A1) = 0.05, P(B|A2) = 0.03, P(B|A3) = 0.01.To answer
the original question, we first find P(B). That may be done
in the following way:
• P(B) = Σi P(B|Ai)P(Ai) = (0.05)(0.2) + (0.03)(0.3) + (0.01)(0.5)
= 0.024.
• Hence 2.4% of the total output of the factory is defective
• We are given that B has occurred, and we want to calculate
the conditional probability of A3. By Bayes' theorem,
• Given that the item is defective, the probability that it was
made by the third machine is only 5/24. Although machine
3 produces half of the total output, it produces a much
smaller fraction of the defective items. Hence the
knowledge that the item selected was defective enables us
to replace the prior probability P(A3) = 1/2 by the smaller
posterior probability P(A3|B) = 5/24.
• 1% of people have a certain genetic defect.
90% of tests for the gene detect the defect.
9.6% of the tests are false positives.
If a person gets a positive test result, what are
the probability they actually have the genetic
defect?
• The first step into solving Bayes theorem
problems is to assign letters to events:
• A = chance of having the faulty gene. That was
given in the question as 1%. That also means
the probability of not having the gene (~A) is
99%.
• X = A positive test result.
• So:
• P(A|X) = Probability of having the gene given a positive test
result.
• P(X|A) = Chance of a positive test result given that the
person actually has the gene. That was given in the
question as 90%.
• p(X|~A) = Chance of a positive test if the person doesn’t
have the gene. That was given in the question as 9.6%
• Now we have all of the information we need to put into the
equation:
P(A|X) = (.9 * .01) / (.9 * .01 + .096 * .99) = 0.0865 (8.65%).
• In a class, 30% have grey eyes, 50% blue eyes
and other 20% have other colours. One day
they play a game together. In the first run,
65% of the grey ones, 82%blue eyes and 50%
with other colours are selected. Now if a child
is selected randomly from the class, and we
know that she was not in the first game, what
is the prob that the child has blue eyes.