Conditional Probability Target Goal: I can find the probability of a conditional event. 5.3b h.w: p 329: odd, 73, 83, 91 – 95 odd

 Warm up: Grade Distributions Consider the two-way table on page 314. Define events E : the grade comes from an EPS course, and L : the grade is lower than a B. Conditional Probability and Independence Find P(L) Find P(E | L) Find P(L | E) Total Total P(L) = 3656 / = P(E | L) = 800 / 3656 = P(L| E) = 800 / 1600 =

The probability that event A occurs if we know for certain that event B will occur is called conditional probability. The conditional probability of A given B is denoted:  P(A/B)  Read as the probability of A given B

Example: Drawing an Ace Deal 4 cards: P(ace) = 4/52 = 1/13 P(ace/you have 1 ace in 4 visible cards) = 3/48 = 1/16

Example: Marital Status of Women  The table shows the marital status of adult women broken down by age.

Define the events  A = the woman is young, ages 18 to 29  B = the woman is married

From Table  A = the woman is young, ages 18 to 29  P(A) = = 22,512/103,870 = 0.217

The probability we chose a woman who is both young and married is:  P(A and B) =  7,842/103,870 = ,870: out of total population

Find the conditional probability she is married given she is young.  P(B/A) =  = 7,842/22,512 = ,512: out of young women

Observation  The probability she is married if we know she is young is much higher (0.38) than if we chose at random (0.075).  Be careful! It is easy to confuse these.

Summary  The probability that a woman is both young and married is,  The product of the probabilities she is young and that she is married given she is young. P(A and B) = P(A) x P(B/A) = 22,512 x 7, ,870 22,512 = she is young she is married given she is young

General Multiplication Rule For Any Two Events The probability any two events occur together is: P(A and B) = P(A) ∙ P(B/A)

Example: Slim Wants Diamonds Slim wants to draw two diamonds in a row. There are 11 cards upturned on the table of which 4 are diamonds. To find Slims probability of drawing two diamonds in a row, first calculate:

P(first card is a diamond) = 9/41 P(2nd card diamond/ 1st card diamond) = 8/40 Slim finds the probabilities by counting cards.

The multiplication rule says that the P(both cards are diamonds) = P(2nd card diamond/ 1st card diamond) = = P(first card is a diamond)x 9/41x 8/40

We can now rearrange the order and find the conditional probability in terms of the unconditional.

 Calculating Conditional Probabilities If we rearrange the terms in the general multiplication rule, we can get a formula forthe conditional probability P ( B | A ). Conditional Probability and Independence P(A ∩ B) = P(A) P(B | A) General Multiplication Rule To find the conditional probability P(B | A), use the formula Conditional Probability Formula P(A ∩ B)P(B | A)P(A)P(A) =

 Example: Who Reads the Newspaper?(Venn diagram) In Section 5.2, we noted that residents of a large apartment complex can be classified based on the events A : reads USA Today and B : reads the New York Times. The Venn Diagram below describes the residents. What is the probability that a randomly selected resident who reads USA Today also reads the New York Times ? Conditional Probability and Independence There is a 12.5% chance that a randomly selected resident who reads USA Today also reads the New York Times.

Example: Marital Status cont. (two way table) Find the conditional probability that a woman is a widow given that she is at least 65 years old. Find: P(widow / at least 65) Need: P(widow and at least 65) and P(at least 65)

What probabilities do we need form the table? P(at least 65) = 18,669/103,870 = P(widow and at least 65) = 8,385/103,870 = 0.081

The conditional probability P(widow/at least 65) =P(widow and at least 65)/ P(at least 65) = 0.081/0.180 = 0.450

Check that this agrees with the result from the “65 and over” column.  P(widow/at least 65)  = 8,385/18,669  = 0.449