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Conditional Probability Target Goal: I can find the probability of a conditional event. 5.3b h.w: p 329: 65 - 69 odd, 73, 83, 91 – 95 odd

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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 3392 2952 3656 10000 Total 6300 1600 2100 P(L) = 3656 / 10000 = 0.3656 P(E | L) = 800 / 3656 = 0.2188 P(L| E) = 800 / 1600 = 0.5000

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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

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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

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

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

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

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The probability we chose a woman who is both young and married is: P(A and B) = 7,842/103,870 = 0.075 103,870: out of total population

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Find the conditional probability she is married given she is young. P(B/A) = = 7,842/22,512 = 0.348 22,512: out of young women

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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.

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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 7,842 103,870 22,512 = 0.075 she is young she is married given she is young

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

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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:

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

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The multiplication rule says that the P(both cards are diamonds) = P(2nd card diamond/ 1st card diamond) = = 0.044 P(first card is a diamond)x 9/41x 8/40

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

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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) =

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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.

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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)

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What probabilities do we need form the table? P(at least 65) = 18,669/103,870 = 0.180 P(widow and at least 65) = 8,385/103,870 = 0.081

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

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

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1. Try to estimate the likelihood of a result by actually observing the random phenomenon many times and calculating the relative frequency of the results.

1. Try to estimate the likelihood of a result by actually observing the random phenomenon many times and calculating the relative frequency of the results.

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