Day 16

More on conditional probability, independence and

total probability formula

Conditional Probability and Independence Example: Roll a fair die twice, the first roll is 6 and the

second roll is also 6. Let A={first roll is 6} and B={second roll is 6} we can see that since two rolls are

independent, P(A ∩ B)=P(A) P(B|A), but P(B|A)=P(B),

therefore, For independent events:

P(A ∩ B) =P(A)P(B)

Independence

Intuitively, if knowing the probability of one event, A, does not help us get the probability of the other event, B, we call the events A and B independent.

P(A|B)=P(A) (NOT P(B) !!!) P(B|A)=P(B) P(A∩B)=P(A)P(B)

Some more exercises on conditional probability According to the survey on longevity,

nowadays, 95% people will live past 20 years and 50% people will live past 65.

Tom just had his twentieth birthday on Sept 22nd, what is his chance of living past 65?

A={past 20} and B={past 65} Here, we want P(B|A)=P(A∩B)/P(A) Think about P(A∩B)…

More on conditional Let A={ it will rain today }, B={ it will

rain tomorrow }. We know that P(A)=0.4, P(B)=0.5 and P(A∩B)=0.3. Given that is rains today, what is the chance that it also rains tomorrow? (greater or less than P(B) ?)

Which probability are we looking for? P(B|A) P(B|A)=P(A∩B)/P(A)=0.3/0.4=0.75

Rain problem contd.

If it does not rain today, what is the chance that it rains tomorrow?

The probability we look for is? P(B|Ac) P(B|Ac)=P(Ac∩B)/P(Ac)

Some equalities

1. P(Ac∩B) + P(A∩B) = P(B)

2. P(Ac|B) + P(A|B) = 1

Keep pushing on the rain problem

Let A={ it rains on Saturday }, B={ I will go to Turkey Run state park}. I know that if it does not rain, I have a 80% chance going, if it rains, I may still go with 20% chance. If the chance of rain on Saturday is 60%, what is my total chance of going?

Total Probability Formula

If the outcome of an event, A, depends on the outcome of another event, B. The chance of A happening is:

P(A)=P(B1)P(A|B1)+ P(B2)P(A|B2)+…+ P(Bk)P(A|Bk), B1, B2, …, Bk are the outcomes of B.

Example Suppose you want to catch an early flight

at Indy airport. You have the following plans: 1. Let your friend drive you to the airport, with

20% chance of missing the flight. 2. Taking the Lafayette Limo with a 25%

chance of missing the flight. 3. Hitch hike yourself with 50% of missing the

flight.If your preference is 40% taking Lafayette Limo,

35% hitch hike and 25% asking your friend to drive, what is your chance of missing the flight.

Another Example

A product is manufactured over three separate machines: A, B and C. Machine A has 4% defective rate, machine B has 2% defective rate and machine C has 6% defective rate. If the percentages of products by machine are: 50%, 30% and 20%, what will be your defective rate?

A summary example

Toss a coin twice. For each toss P{H}=0.7 and P{T}=0.3. Find:

1. P{two heads}; 2. P{two tails}

Summary example (contd)

3. P{one head and one tail}

Yet another example Someone is shooting at a target. If it is

windy, he has a 40% chance hitting the target. If there is no wind, his chance is 70%.

A. If there is a 12% chance of being windy, what is his chance of hitting the target?

B. If he hits the target, what is the chance of it being windy???

Question B can only be solved using a new technique Bayes theorem:

Bayes theorem deals with another type of question.

Think about conditional probability: it answers the question: if A happens then what is the chance for B to happen? ( A is a condition for B)

Bayes theorem answers another question: If B happens, what is the chance of A happening. (A is still a condition for B )

Formula Think about total probability formula

(two events case) P(A)=P(B1)P(A|B1)+P(B2)P(A|B2)

Now we want to know P(B1|A)

Bayes’ Theorem

P(B1|A)= P(B1)P(A|B1) / [P(B1)P(A|B1)+P(B2)P(A|B2)]= P(B1)P(A|B1) / P(A)

P(B2|A)= P(B2)P(A|B2) / [P(B1)P(A|B1)+P(B2)P(A|B2)]

= P(B2)P(A|B2) / P(A)