Chris morgan math g160 csmorgan@purdue edu january 13 2012 lecture 3
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Chris Morgan, MATH G160 [email protected] January 13, 2012 Lecture 3. Chapter 4.4: Conditional Probability, Multiplication Rule, Law of Total Probability. Conditional Probability. • The probability an event occurs under the condition that another event occurred

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Chris Morgan, MATH G160 [email protected] January 13, 2012 Lecture 3

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Chris morgan math g160 csmorgan@purdue edu january 13 2012 lecture 3

Chris Morgan, MATH G160

[email protected]

January 13, 2012

Lecture 3

Chapter 4.4:

Conditional Probability, Multiplication Rule,

Law of Total Probability


Chris morgan math g160 csmorgan purdue january 13 2012 lecture 3

Conditional Probability

• The probability an event occurs under the condition that another event occurred

- denoted: P(A|B)

- “The probability of A given B”

P (A | B)

want to find given already know happened


Chris morgan math g160 csmorgan purdue january 13 2012 lecture 3

Conditional Probability


Chris morgan math g160 csmorgan purdue january 13 2012 lecture 3

Conditional Probability


Chris morgan math g160 csmorgan purdue january 13 2012 lecture 3

Conditional Probability

Example

In a certain population:

- The probability a person lives to be 80 is 75%

- The probability a person lives to be 90 is 63%

Find the probability an 80 year old will live to be 90:

P(live to 90 | live to 80) = P(A intersect B) / P(B)

= (0.63) / (0.75)

= 0.84


Chris morgan math g160 csmorgan purdue january 13 2012 lecture 3

Conditional Probability

Possible ways to toss two fair dice:


Chris morgan math g160 csmorgan purdue january 13 2012 lecture 3

Conditional Probability

Roll two fair six-sided dice:

P(sum 7 | first die is a 4) = P(second die is a 3) = 1/6

or:


Chris morgan math g160 csmorgan purdue january 13 2012 lecture 3

Conditional Probability

You and your roommate have 9 frozen pizzas: 4 veggie and 5 meat lovers. You randomly select a pizza out of the freezer have it for dinner tonight. Tomorrow, your roommate goes to the store and get 3 more pizzas of which ever kind you ate today.

P(eat veggie today) = 4/9

P(eat veggie tomorrow | ate veggie today) = ?

since I ate veggie today, I know my roommate bought three more veggie pizzas, so I now have (4-1+3) 6 veggie pizzas in the freezer, and the probability I eat one of those 6 veggie pizzas out of the total eleven in my freezer is 6/11 = 0.545


Chris morgan math g160 csmorgan purdue january 13 2012 lecture 3

Conditional Probability Example

At a fair a vendor has 25 helium balloons on strings: 10 balloons are yellow, 8 are red, and 7 are green. A balloon is selected at random and sold. Given that the balloon sold is yellow, what is the probability that the next balloon selected at random is also yellow?

Event A = next random balloon is yellow

Event B = sold yellow balloon

P (A|B) = 9 / 24


Chris morgan math g160 csmorgan purdue january 13 2012 lecture 3

General Multiplication Rule

With two events:

With three events:

With n events:


Chris morgan math g160 csmorgan purdue january 13 2012 lecture 3

Law of Total Probability

Two events A and B are mutually exclusive if:

Example:

P(fair die shows multiple of 3) = P(shows 3) + P(shows 6)


Chris morgan math g160 csmorgan purdue january 13 2012 lecture 3

Law of Total Probability

The generalization to n mutually exclusive events is the law of total probability:


Chris morgan math g160 csmorgan purdue january 13 2012 lecture 3

Law of Total Probability

The collection of events is called a partition, mutually exclusive and exhaustive events, for the sample space omega if the events are pair-wise disjoint and they exhaust omega. In other words:

whenever k ≠ m and


Chris morgan math g160 csmorgan purdue january 13 2012 lecture 3

Law of Total Probability

For example, suppose {B1, B2, B3, B4} form a partition for omega and we want to calculate the probability of event E:

We can use the partition to decompose E into four disjoint pieces:

B1

B2

B3

B4

E


Chris morgan math g160 csmorgan purdue january 13 2012 lecture 3

Total Probability example

Acme Consumer Goods sells three brands of washing machines, Brand I, Brand II and Brand III. 30% of the machines they sell are Brand I, 50% are Brand II, and 20% are Brand III. Based on past experience Acme executives know that the purchasers of Brand I machines will need service repairs with probability .2, Brand II machines with probability .15, and Brand III machines with probability .25. Find the probability a customer will need service repairs on washing machine they purchased from Acme.


Chris morgan math g160 csmorgan purdue january 13 2012 lecture 3

Example 10

A grade school boy has 5 blue and 4 white marbles in his left pocket and 4 blue and 5 white marbles in his right pocket. If he transfers one marble at random from his left pocket to his right pocket, what is the probability of his then drawing a blue marble from his right pocket?


Chris morgan math g160 csmorgan purdue january 13 2012 lecture 3

Example 11

Suppose that there are 14 songs on a MP3 player and you like 8 of them. When using the random song selection option each of the 14 songs is played once in a random order. Find the probability that among the first two songs that are played.

a) You like both of them.

b) You like neither of them.

c) You like exactly one of them.


Chris morgan math g160 csmorgan purdue january 13 2012 lecture 3

Example 12

A drawer contains four black, six brown, and eight olive socks. Two socks are selected at random from the drawer.

a) Compute the probability that both socks are the same color.

b) Compute the probability that both socks are olive if it is known that they are the same color.


Chris morgan math g160 csmorgan purdue january 13 2012 lecture 3

Example 12

98% of all babies survive delivery. However, 15% of all births involve Cesarean sections, and when a C section is performed the baby survives 96% of the time. If a randomly chosen pregnant woman does not have a C section, what is the probability that her baby survives?


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