Download General addition rule

Probability, contd
General Addition Rule
General Multiplication Rule
Conditional Probability
Learning Objectives
By the end of this lecture, you should be able to:
– Apply the general addition rule and the general multiplication
rule.
– Describe what is meant by the term ‘general’ in the general
addition rule and general multiplication rule.
– Describe and apply the conditional probability rule.
Example
Example: What is the probability of randomly drawing either an ace or a
heart from a deck of 52 playing cards?
Note that you can not use our original version of the addition rule here since the
events are not disjoint. (It is possible for the card to be both an Ace and a Heart).
Example: If rolling a single die, determine the probability of
rolling an even number, or a number greater than 2.
P(Even or >2) = ?
P(Rolled an Even)
P(Rolled a number >2)
P(Rolled an Even OR >2)
= 3/6
= 4/6
= P(Rolled an Even) + P(Rolled a number >2)
=
3/6
+
4/6
=
7/6 ?!?!
No, because in this case, there are some non-disjoint outcomes (in this case, two). Two of the
outcomes: 4 and 6 are even and also greater than two. Therefore, these two outcomes were counted
twice.
However, we CAN apply the so-called general addition rule which works on both disjoint events, and
ALSO on NON-disjoint events…
General addition rule
P(A or B) = P(A) + P(B) – P(A and B)
Why is it called the “general” rule?
“General” means can be used on BOTH disjoint and non-disjoint events!
Example: If rolling a single die, determine the probability of
rolling an even number, or a number greater than 2.
P(A or B) = P(A) + P(B) – P(A and B)
P(Even or >2) = ?
P(Outcome is an Even)
= 3/6
P(Outcome is a number >2)
= 4/6
P(Outcome is an Even AND >2) = 2/6
Applying General Addition Rule:
P(A or B)
=
P(A)
=
P(Even)
=
3/6
= 5/6
+
+
P(B)
P(>2)
+
4/6
– P(A and B)
– P(Even and >2)
–
2/6
Why do we call it the “general” addition rule?
• Because it applies to ANY addition events. You can use it for BOTH disjoint
events and non-disjoint events!
• Why does it also work for disjoint events?
– Recall that if 2 events are disjoint, this means that the two events are
mutually exclusive. In other words, if 1 event is true, the other must be false.
– Therefore, P(A and B), i.e. the probability of both events being true will
always equal 0.
So: P(A or B) = P(A) + P(B) + P(A and B) .
However, if the events are disjoint, then P(A and B) is 0,
Therefore: P(A or B) = P(A) + P(B) + 0 (i.e. This is our addition rule for disjoint
events)
Here is an example of applying the general rule to a disjoint event…
Example: What is the probability of randomly drawing
P(A or B) = P(A) + P(B) – P(A and B)
either an Ace or a 7 from a deck of 52 playing cards?
•
P(Card is an Ace)
 4/52
•
P(Card is a 7)
 4/52
•
P(Card is an Ace AND a 7)
0
P(Draw an Ace OR Draw a 7) ?
= P(Ace) + P(7) – P(Ace and 7)
= 4/52 + 4/52 –
0/52)
= 8/52
Example: What is the probability of randomly drawing
either an ace or a heart from a deck of 52 playing cards?
P(A or B) = P(A) + P(B) – P(A and B)
•
P(Ace)
 4/52
•
P(Heart)
 13/52
•
If we simply added them, we would get 17/52. This is NOT the correct result!
•
P(Ace and Heart)
•
There is one NON-disjoint event present. Notice how the Ace of Hearts has been counted twice. Therefore we must subtract this doubled item. So
 1/52
the correct answer is: (4/52 + 13/52 – 1/52) = 16/52.
General addition rule
General addition rule for any two events A and B:
P(A or B) = P(A) + P(B) – P(A and B)
What is the probability of randomly drawing either an ace or a heart from a deck of 52 playing cards?
Answer: There are 4 aces in the pack and 13 hearts. However, 1 card is both an ace and a heart. If you
simply added the two probabilities separately, you would end up counting that same card twice.
The general addition rule tells us that if some of the outcomes are disjoint, then we will overcount those
disjoint outcomes – an additional time for each disjoint event.
Therefore, we need to subtract those overlaps. In this problem, there is exactly one disjoint event.
Thus: P(ace or heart)
= P(ace) + P(heart) – P(ace and heart)
= 4/52 (the 4 aces) + 13/52 (the 13 hearts) - 1/52 (the Ace of Hearts)
= 16/52
* Incidentally, you may be tempted to try to use the multiplication rule to calculate P(Ace and Heart). However, these
events are not independent, so you can’t use your multiplication rule just yet. For this reason, simply accept my word
(or determine intuitively) that the chance of P(Ace and Heart) is exactly 1/52.
Example: In a group of 100 athletes, 30% are basketball players. In that same group, 25% are over 6feet tall. What is P(BB or >6 feet)?
Answer:
– P(Basketball Player) = 0.3,
– P(>6 feet tall) = 0.25
If we count 0.3 + 0.25, we will get an artificially high number. This is because in the basketball group we
will count several people who are also 6 feet. Similarly, among the 6-footers, we will count several
basketball players. Therefore, we will have counted those people TWICE.
To get an accurate result, we have to subtract the number of outcomes that we counted twice. Again,
who got counted twice? Answer: Those people who are both basketball players and 6 feet:
P(BB and >6 feet) = P(BB) + P(6’) – P(BB and 6’)
Note that in this question, we do not have enough information to calculate the probability since we are
not told how many people in the 100 are both basketball players and 6 feet tall. This would be one of
those (horrible) “not enough information to answer the problem” questions. However, in the real
world, people often make mistakes as a result of coming up with “answers” in spite of having
incomplete data. You need to be able to recognize when you are in this situation.
More than one non-disjoint event
Example: What is the probability that a card from a deck is either a
King or a Queen or a Diamond?
P(King) + P(Queen) + P(Diamond) is NOT correct since there are non-disjoint events that
will be overcounted.
Non disjoint events: King of Diamonds and Queen of Diamonds
To solve this question, we count all the outcomes, and then subtract all outcomes
that have overlapped. I.e. All non-disjoint outcomes.
= P(King) + P(Queen) + P(Diamond) – P(King and Diamond) – P(Queen and
Diamond)
= 4/52 +
4/52 +
13/52 –
1/52
–
1/52
= 19 /52
Independence revisited
Suppose you tend to spend all your free weekends in Seattle (if the boss gives you the
weekend off) and you want to determine whether or not you need to dig out your
raincoat.
• Your boss gives you roughly 30% of your weekends off, so P(you will to go Seattle) =
0.3.
• It rains roughly 10% of days: P(Rain) = 0.1
Using our original multiplication rule, we might be temped to say: P(in Seattle AND Rain)
= 0.3 * 0.1 = 0.03.
Thoughts?
These events P(in Seattle) and P(Rain) are NOT independent: The proabability that it is
going to rain will indeed be affected by your being in Seattle as opposed to, say, being in
Death Valley, California.
Multiplication Rule revisited
Suppose you tend to spend all your free weekends in Seattle (if the boss gives
you the weekend off) and you want to determine whether or not you need to
dig out your raincoat.
P(in Seattle) = 0.3
P(Rain given that we are in Seattle) = 0.4
KeyPoint: We needed to adjust our probability of rain to account being in
Seattle. When we said P(Rain) = 0.1 a moment ago, we were taking some kind
of national average. However, P(Rain given that we are in Seattle) is about 0.4.
The moment you find yourself saying “given that” (or something similar), you
are talking about a conditional probability. In other words, you are
acknowledging that the probability you are interested in may be affected by
something else. In thjs case, you are acknowledging that the probability of
Rain is affected by being in Seattle.
Key Point
The moment events are NOT independent, you must recognize
that somewhere in your probability calculation, you will need to
apply a conditional probability.
– Probability of drawing an Ace and then drawing another Ace. P(2nd
card is an Ace) changes depending on whether or not the first card
was an Ace. These events are NOT independent, and therefore, you
must include a conditional probability in your calculations.
General multiplication rule (“And”)
When dealing with events that are not independent, we need to look at
our ‘conditional’ event and account for the possible change in
probability.
– Recall that if A and B are independent, then P(A and B) = P(A) * P(B)
– However, if P(B) changes based on whether or not A has occurred, then we
are saying that the events are not independent.
– Therefore, rather than simply saying P(B), we must adjust it to say P(B given
that A has occurred).
– There is a special notation for this: P(B | A).
P(A and B) = P(A) * P(B|A)
This is called the general multiplication rule. That is, this is aversion of
the multiplication rule that is not limited to independent events.
P(A and B) = P(A) * P(B|A)
Example: What is the probability of randomly drawing a card from the deck that is an Ace AND a Heart?
P(ace and heart) = P(ace) * P(heart | ace)
P(Ace)
= (4/52)
P(Heart | Ace)
= (1/4)
 Take a moment and think about this! We are limiting the situation to Aces only!!
 Probability of a Heart GIVEN that we are looking at Aces = 1/4
Answer: P(ace) * P(heart | ace)
= (4/52) * (1/4)
= 1/52
Why do we call it the “general” multiplication rule?
• Same story as with the “general” addition rule. That is, this
rule applies to ANY multiplication events – BOTH
independent and non-independent.
Why does it also work for independent events?
– Recall that if two events are independent, this means that P(B) is NOT
affected by P(A).
– That is, P(B | A) = P(B).
– Our general rule states: P(A and B) = P(A) * P(B | A)
• If our events are independent, then P(B | A ) = P(B)
• So: P(A and B) = P(A) * P(B)
P(A|B) is not the same as P(B|A)
• P(Heart | Ace) is not the same thing as P(Ace | Heart).
– P(Heart | Ace) = 1/4
– P(Ace | Heart) = 1/13
• P(Rain | Seattle) is not the same thing as P(Seattle | Rain)
• Key Point: Don’t interchange them! Keep track of which is
which!
Calculating the Conditional probability
Conditional probabilities reflect how the probability of an event can change
if we know that some other event has occurred.
– Example: The probability that a cloudy day will result in rain is different if you
live in Las Vegas than if you live in Seattle.
– Every single day, our brains calculate conditional probabilities, updating our
“degree of belief” with each new piece of evidence.
Notation: The conditional probability
of event B “given” event A is:
Note: We assume that P(A) ≠ 0
P( A and B)
P( B | A) 
P( A)
Spoken as: “Probability of B given A”
Example
• About 30% of women with early breast cancer will experience metastases,
a spread of cancer from its initial location. A new genetic test was
developed to help identify those women with early breast cancer who will
later develop metastases. A study of women with early breast cancer who
took this new genetic test finds that 27% had a positive test and later
developed metastases. What is the new test’s ability to identify women
who will develop metastases?
• Answer:
Restated: “What is the probability that a women gets a positive test, given that she later
develops metastases?” In other words: P(Test is Positive | Metastases)?
P(Test + | Mets )
= P(Test + AND Metastases)
=
0.27
=
/ P(Metastases)
/
0.3
0.9
Example
• Slim is playing poker and holds 3 diamonds. He hopes to draw
two diamonds in a row in order to get a flush (all cards of the
same suit). Of all the cards showing (those in Slim’s hand and
those upturned on the table), he sees 11 cards. 4 of those 11
cards are diamonds. So 9 of the 41 remaining unseen cards
must be diamonds. What is the probability that Slim will draw
his two diamonds?
– P(first card is a diamond) = 9/41
– P(second card diamond | given first card diamond) = 8/40
– 9/41 * 8/40 = 0.044
Conditional probability can be applied to more than two events
• What is the probability that a person chosen at random off
the street is: Italian, Male and plays the lute?
–
–
–
–
A = probability of being Italian
B = male
C = plays lute
P(A and B and C) = P(A) * P(B | A) * P(C | A and B)
• Key Point: It can get a bit unwieldy, but if you have the data, it
is possible to do these calculations. Don’t panic though: In this
course, I will never ask you to do more than one conditional
probability.