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Conditional Probabilities Pr (E|F) = Pr(E∩F)/Pr(F) Pr (F|E) = Pr(E∩F)/Pr(E)

Conditional Probabilities Pr (E|F) = Pr(E∩F)/Pr(F) Pr (F|E) = Pr(E∩F)/Pr(E)

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Page 1: Conditional Probabilities Pr (E|F) = Pr(E∩F)/Pr(F) Pr (F|E) = Pr(E∩F)/Pr(E)

Conditional Probabilities

Pr (E|F) = Pr(E∩F)/Pr(F)

Pr (F|E) = Pr(E∩F)/Pr(E)

Page 2: Conditional Probabilities Pr (E|F) = Pr(E∩F)/Pr(F) Pr (F|E) = Pr(E∩F)/Pr(E)

Example 1

Four-letter word: 2 A’s & 2 B’s

Pr(2 A’s together | Last letter B) = ?

Page 3: Conditional Probabilities Pr (E|F) = Pr(E∩F)/Pr(F) Pr (F|E) = Pr(E∩F)/Pr(E)

Multiplicative Laws

Pr(E∩F) = Pr(F) Pr (E|F)

Pr(E∩F) = Pr(E) Pr (F|E)

Page 4: Conditional Probabilities Pr (E|F) = Pr(E∩F)/Pr(F) Pr (F|E) = Pr(E∩F)/Pr(E)

Example 2

A box contains 5 & 10 balls

Two balls picked without replacement.

Describe the outcomes & probabilities:

Page 5: Conditional Probabilities Pr (E|F) = Pr(E∩F)/Pr(F) Pr (F|E) = Pr(E∩F)/Pr(E)

Example 3

A bag contains 5 balls, 3 are red and 2 are yellow. Three balls are drawn without replacement. What is the probability of drawing at least two red balls?

Page 6: Conditional Probabilities Pr (E|F) = Pr(E∩F)/Pr(F) Pr (F|E) = Pr(E∩F)/Pr(E)

Total Probability Law

Pr(E) = Pr(F) Pr (E|F) + Pr(Fc) Pr (E|Fc)

Page 7: Conditional Probabilities Pr (E|F) = Pr(E∩F)/Pr(F) Pr (F|E) = Pr(E∩F)/Pr(E)

Bayes Theorem

Pr(F|E) = Pr (F) Pr (E | F)/{Pr(F) Pr (E|F) + Pr(Fc) Pr (E|Fc)}

Page 8: Conditional Probabilities Pr (E|F) = Pr(E∩F)/Pr(F) Pr (F|E) = Pr(E∩F)/Pr(E)

Example 4

A school has 60% Boys & 40% Girls.

Of the boys, 80% have activity tickets.

Of the girls, 75% have activity tickets

Pr(belongs to a boy | ticket found) = ?

Pr(belongs to a girl | ticket found) = ?

Page 9: Conditional Probabilities Pr (E|F) = Pr(E∩F)/Pr(F) Pr (F|E) = Pr(E∩F)/Pr(E)

Example 5

Suppose that colored balls are distributed in two indistinguishable boxes as follows:

Box 1 Box 2Red 2 4Yellow 3 1

A box is selected at random from which a ball is selected at random and it is observed to be red. What is the probability that box 2 was selected? Box 1 selected?

Page 10: Conditional Probabilities Pr (E|F) = Pr(E∩F)/Pr(F) Pr (F|E) = Pr(E∩F)/Pr(E)

Example 6

As accounts manager in your company, you classify 75% of your customers as “good credit” and the rest as “risky credit” depending on their credit rating. Customers in the “risky” category allow their accounts to go overdue 50% of the time on average, whereas those in the “good” category allow their accounts to become overdue only 10% of the time. What percentage of overdue accounts are held by customers in the “risky credit” category?

Page 11: Conditional Probabilities Pr (E|F) = Pr(E∩F)/Pr(F) Pr (F|E) = Pr(E∩F)/Pr(E)

Independence of Events

Pr(E∩F) = Pr(E) Pr (F)

Pr(E|F) = Pr(E)

Pr(F|E) = Pr(F)

Page 12: Conditional Probabilities Pr (E|F) = Pr(E∩F)/Pr(F) Pr (F|E) = Pr(E∩F)/Pr(E)

Example 7

Two coins tossed:

E = “Head on first coin”

F = “Coins fall alike”

Are E and F independent?

Page 13: Conditional Probabilities Pr (E|F) = Pr(E∩F)/Pr(F) Pr (F|E) = Pr(E∩F)/Pr(E)

Example 8

A town has two fire engines operating independently. The probability that a specific fire engine is available when need is 0.99. What is the probability that neither is available when needed? What is the probability that a fire engine is available when needed?

Page 14: Conditional Probabilities Pr (E|F) = Pr(E∩F)/Pr(F) Pr (F|E) = Pr(E∩F)/Pr(E)

Example 9

You throw two fair dice, one green and one red. Decide which of the following pairs of events are indpendent:

A) Sum is 5 & Red die shows 2

B) Sum is 5 & Red die shows even

C) Sum is 5 & Sum is 4

D) Sum is even & Red die shows even

Page 15: Conditional Probabilities Pr (E|F) = Pr(E∩F)/Pr(F) Pr (F|E) = Pr(E∩F)/Pr(E)
Page 16: Conditional Probabilities Pr (E|F) = Pr(E∩F)/Pr(F) Pr (F|E) = Pr(E∩F)/Pr(E)

Example 10

Two cards are drawn from a bridge deck. What is the probability that the second card drawn is red?

Page 17: Conditional Probabilities Pr (E|F) = Pr(E∩F)/Pr(F) Pr (F|E) = Pr(E∩F)/Pr(E)

Example 11

What is the probability that a family of two children has:

(a) two boys given that it has at least one boy?

(b) two boys given that the first child is a boy?

Page 18: Conditional Probabilities Pr (E|F) = Pr(E∩F)/Pr(F) Pr (F|E) = Pr(E∩F)/Pr(E)

Example 12

Assume that E and F are two events with positive probabilities. Show that if Pr (E | F) = Pr (E) then Pr (F | E) = Pr (F).

Page 19: Conditional Probabilities Pr (E|F) = Pr(E∩F)/Pr(F) Pr (F|E) = Pr(E∩F)/Pr(E)

Example 13

Prove that for any three events A, B, C, each having positive probability, Pr (A ∩ B ∩ C) = Pr (A) Pr (B | A) Pr (C | A ∩ B).

Page 20: Conditional Probabilities Pr (E|F) = Pr(E∩F)/Pr(F) Pr (F|E) = Pr(E∩F)/Pr(E)

Example 14

Suppose that A and B are events such that Pr (A | B) = Pr (B | A) and Pr(A U B) = 1 and Pr(A ∩ B) > 0. Prove that Pr (A) > 1/2.

Page 21: Conditional Probabilities Pr (E|F) = Pr(E∩F)/Pr(F) Pr (F|E) = Pr(E∩F)/Pr(E)

Example 15

We are given three coins: one has heads in both faces, the second has tails in both faces, and the third has a head in one face and a tail in the other. We choose a coin at random, toss it, and it comes heads. What is the probability that the opposite face is tails?

Page 22: Conditional Probabilities Pr (E|F) = Pr(E∩F)/Pr(F) Pr (F|E) = Pr(E∩F)/Pr(E)

Example 16

A batch of one hundred items is inspected by testing two randomly selected items. If one of the two is defective, the batch is rejected. What is the probability that the batch is accepted if it contains five defectives?

Page 23: Conditional Probabilities Pr (E|F) = Pr(E∩F)/Pr(F) Pr (F|E) = Pr(E∩F)/Pr(E)

Example 17

Let A and B be events. Show that Pr (A ∩ B |B) = Pr (A|B), assuming that Pr (B) > 0.

Page 24: Conditional Probabilities Pr (E|F) = Pr(E∩F)/Pr(F) Pr (F|E) = Pr(E∩F)/Pr(E)

Example 18

The king has only one sibling. What is the probability that the sibling is male?