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7.5 Conditional Probability; Independent Events This presentation is copyright by Paul Hendrick © , Paul Hendrick All rights reserved

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7.5 Conditional Probability; Independent Events To figure the probability of an event E, weve been using the sample space S and the formula P(E) = n(E) / n(S) Another example: –Three fair coins are tossed (or one fair coin is tossed three times). –Whats the probability of having 2 heads (on 3 tosses) ? –In symbols, P(2H) = ? –P(2H) = 3/8 (How? Lets see…)

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7.5 Conditional Probability; Independent Events Recall, P(E) = n(E) / n(S) –We need to figure n(S) and n(E), (in that order -- recommended!) In our example, there are t hree tosses, so –S= {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT} Whats the probability of having 2 heads (on 3 tosses) ? –So E = 2H = {HHT, HTH, THH} Then P(E) = P(2H) = n(2H) / n(S) = 3/8 Note that above, 2 means exactly 2, and HHH doesnt count!

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7.5 Conditional Probability; Independent Events Lets modify the above problem : –Three fair coins are tossed (or one fair coin is tossed three times). –Whats the probability of having 2 heads (on 3 tosses), given that the first toss was a TAIL ? –In symbols, P(2H|T 1 ) = ? The | means given that and the superscript 1 means on the first toss –P(2H|T 1 ) = 1/4 (How? Lets see…)

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7.5 Conditional Probability; Independent Events Again, it would seem that S= {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT} But … we are told that the first toss was a tail! So, not all of the above were really possible! So we modify the sample space to reflect what could have possibly happened: S ~ = {THH, THT, TTH, TTT} = T 1 Of these, our event E would be as follows: E = 2H = {THH} Then, P(2H|T 1 ) = 1/4 –Note that the sample space was restricted (or reduced)

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7.5 Conditional Probability; Independent Events Restricted sample space This is called Conditional probability –P(A|B) – whats the probability that A occurs, on condition that B has happened (or will have happened) ? –formula: P(A|B) = n( A B ) / n(B) (all equally- likely outcomes!) –or P(A|B) = P(A B ) / P(B) (in general). of course, denominator cannot equal 0, else result is UNDEFINED –In which case, the conditional probability has no meaning. NOTE: the right hand side is always P(both) / P(given)

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7.5 Conditional Probability; Independent Events –Dont confuse P(A|B) with P(B|A) -- they generally are DIFFERENT VALUES P(A|B) = P(A B ) / P(B) P(B|A) = P(A B ) / P(A) are equal only if P(A) = P(B) and both Back to the 3-coin example: –P(T 1 |2H) = ? –S ~ = {HHT, HTH, THH} = 2H –E = T 1 = {THH} –So, P(T 1 |2H) = 1/3 –But P(2H|T 1 ) = 1/4 (while ago!)

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7.5 Conditional Probability; Independent Events We saw listing out the elements in the sample space and using formulas to do problems, above. Or we can use visual aids to help us work with probabilities or conditional probabilities. These visual aids include –Venn diagrams –Tables –Tree diagrams –Probability tree diagrams Well look at using Venn diagrams first

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7.5 Conditional Probability; Independent Events Venn diagrams may help illustrate basic probabilities and be handy for working with conditional probabilities, if done carefully Such a diagram may show either counts or probabilities. If counts are used, the sample space must be uniform! Lets look at an example of using a Venn diagram in a conditional probability setting:

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7.5 Conditional Probability; Independent Events A composition course has 25 students of whom 14 are male and 9 are sophomores. In addition, 5 of the sophomores are men. A student is chosen at random. Find the probability that the student –(a) is a sophomore given that the student is a man. –(b) is a woman given that the student is a sophomore. Note that counts are given, but its a uniform sample space, so were OK. –Why do we know its uniform? [on board]

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7.5 Conditional Probability; Independent Events What if we had counts on freshmen, sophomores, and juniors in the class category, as well as on gender? How well could we put this into our Venn diagram?? Tables will help us past this problem. –But first another Venn diagram problem.

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7.5 Conditional Probability; Independent Events Let E and F be events with P(E) =.35, P(F) =.67, and P(E F) =.24. Find P(E|F ). –Note that probabilities are given this time [on board] –Note that above we DIDNT HAVE TO USE DIAGRAMS (we could just use formulas) … –…but they certainly make clearer whats going on –that is, they help us visualize the situation

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7.5 Conditional Probability; Independent Events Tables can be viewed as similar to Venn diagrams, except more versatile. Such a table may show either counts or probabilities. If counts are used, the sample space must be uniform! Lets look at an example of using a table in a conditional probability setting:

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7.5 Conditional Probability; Independent Events –Problem 7.5 # 46. The following table, based on data from the Centers for Disease Control, gives the number of new cases of the AIDS virus for men and women in the United States in 1998 by method of transmission. –Find (a) the probability that a male resident who was newly diag- nosed with AIDS in 1998 contracted it via homosexual contact. Method of transmn.MaleFemaleTotal Homosexual contact8,3880 Heterosexual contact1,1461,8062,952 Intravenous drug use3,6521,5415,193 Other5,2372,0547,291 Total18,4235,40123,824

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7.5 Conditional Probability; Independent Events –Problem 7.5 # 46 (a). Find the probability that a male resident who was newly diagnosed with AIDS in 1998 contracted it via homosexual contact. –The table represents ALL the individual residents. –We are told that the resident is male. This is additional information. Its the given. Then we are asked about the homosexual contact, as a transmission method. Method of transmn.MaleFemaleTotal Homosexual contact8,3880 Heterosexual contact1,1461,8062,952 Intravenous drug use3,6521,5415,193 Other5,2372,0547,291 Total18,4235,40123,824

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7.5 Conditional Probability; Independent Events –Our P(A|B) = n(A B) / n(B) becomes –P(Homosexual contact | Male) or –P(Homos. c.| M) = n(Homos. c. M) / n(M) = 8388 / Method of transmn.MaleFemaleTotal Homosexual contact8,3880 Heterosexual contact1,1461,8062,952 Intravenous drug use3,6521,5415,193 Other5,2372,0547,291 Total18,4235,40123, or 45.53%

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7.5 Conditional Probability; Independent Events –Problem 7.5 # 46. The following table, based on data from the Centers for Disease Control, gives the number of new cases of the AIDS virus for men and women in the United States in 1998 by method of transmission. –Find (b) the probability that a female resident who was newly diagnosed with AIDS in 1998 contracted it via intravenous drug use. Method of transmn.MaleFemaleTotal Homosexual contact8,3880 Heterosexual contact1,1461,8062,952 Intravenous drug use3,6521,5415,193 Other5,2372,0547,291 Total18,4235,40123,824

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7.5 Conditional Probability; Independent Events –Problem 7.5 # 46 (b) the probability that a female resident who was newly diagnosed with AIDS in 1998 contracted it via intravenous drug use. –The table represents ALL the individual residents. –We are told that the resident is female. This is additional information. Its the given. Then we are asked about the intravenous drug use, as a transmission method. Method of transmn.MaleFemaleTotal Homosexual contact8,3880 Heterosexual contact1,1461,8062,952 Intravenous drug use3,6521,5415,193 Other5,2372,0547,291 Total18,4235,40123,824

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7.5 Conditional Probability; Independent Events –Our P(A|B) = n(A B) / n(B) becomes –P(intravenous drug use | Female) or –P(IV drug | F) = n ( IV drug F) / n(F) = 1541 / 5401 Method of transmn.MaleFemaleTotal Homosexual contact8,3880 Heterosexual contact1,1461,8062,952 Intravenous drug use3,6521,5415,193 Other5,2372,0547,291 Total18,4235,40123, or 28.53%

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7.5 Conditional Probability; Independent Events –Tree diagrams are handy for working with conditional probabilities and especially for illustrating such –And in particular, probability tree diagrams, with probabilities written on the branches, and then the path probabilities calculated and written for each of the paths, are extremely helpful –Just as above, we used a formula-based approach first, then illustrated with a Venn diagram or table –Here well do the formula first, then a probability tree diagram to illustrate the idea. –First, lets further develop the formulae.

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7.5 Conditional Probability; Independent Events –Weve had the formulae P(A|B) = P(A B ) / P(B) P(B|A) = P(A B ) / P(A) –If you solve each of these equations for the numerator on the right, you get Product Rule of Probability P(A B ) = P(A) P(B|A) P(A B ) = P(B) P(A|B), also

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7.5 Conditional Probability; Independent Events –Product Rule of Probability The probability of both of two events occurring is the product of the probability of the first event occurring times the conditional probability of the second event occurring given that the first event has occurred. P(A B ) = P(A) P(B|A) P(A B ) = P(B) P(A|B), also The conditional probability is key! This idea is called the Multiplication Principle. –Remember, you are not multiplying the probabilities of the two separate events, but rather the probability(ies) that follow(s) the first must be conditional.

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7.5 Conditional Probability; Independent Events P( A B C ) = P(A) P(B|A) P(C|A B) –uses Prob of A, –then Prob of B, but assuming that A has happened! –then Prob of C, but assuming that both A and B have happened! –The conditional probability is key! Could also be P( A B C ) = P(B) P(C|B) P(A|B C) –the order of the events DOES NOT MATTER in the calculation

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7.5 Conditional Probability; Independent Events Examples: (problems #8,10,12,14 in 7.5 in our textbook) –Two cards are drawn without replacement from an ordinary deck. –Find the following probabilities: #8) The second is black, given that the first is a spade #10) The second is an ace, given that the first is not an ace #12) An ace and a 4 are drawn #14) Two hearts are drawn –Note the without replacement in the description. Means the second card is conditional (depends on first card)!

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7.5 Conditional Probability; Independent Events –#8) Two cards are drawn without replacement from an ordinary deck. Find the probability that the second is black, given that the first is a spade. –In symbols, find P(B 2 |S 1 ) where B 2 represents black on the second card, and S 1 represents a spade on the first card –Think what this question means, and solve, keeping in mind what has happened (use a reduced sample space) : A spade is (was, will be) drawn first, leaving 12 spades and 39 other cards, 51 total; One of these is drawn; to be a success, it must be a black card, i.e., one of the 12 spades or 13 clubs, 25 total. The probability therefore is 25/51. P(B 2 |S 1 ) = 25/51

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7.5 Conditional Probability; Independent Events –#10) Two cards are drawn without replacement from an ordinary deck. Find the probability that the second is an ace, given that the first is not an ace. –In symbols, find P(A 2 |A 1 ) where A 2 represents ace on the second card, and A 1 represents not an ace on the first card –Think what this question means, and solve, keeping in mind what has happened (use a reduced sample space) : Not an ace is (was, will be) drawn first, leaving 47 non-aces and 4 aces, 51 total; One of these is drawn; to be a success, it must be an ace, i.e., one of the 4 aces. The probability therefore is 4/51. P(A 2 |A 1 ) = 4/51

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7.5 Conditional Probability; Independent Events –NOTE: To solve the previous two problems, we didnt use anything high-powered, formulas or such! –All we did have to do was (1) UNDERSTAND THE EXPERIMENT (2) UNDERSTAND THE QUESTION (3) KNOW THAT PROBABILITY CAN BE ANSWERED IN TERMS OF COUNTING (in many situations) and (4) DO SOME SIMPLE COUNTING –Many probability problems, especially conditional probability problems, can be worked within these simple ideas.

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7.5 Conditional Probability; Independent Events –#14) Two cards are drawn without replacement from an ordinary deck. Find the probability that two hearts are drawn. –In symbols, find P(2H) where H represents any heart, so 2H represents two hearts P(2H) = P(H 1 H 2 ) = P(H 1 ) P(H 2 |H 1 ) = 13/52 12/51 = 156/2652 = 1 /

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7.5 Conditional Probability; Independent Events –#12) Two cards are drawn without replacement from an ordinary deck. Find the probability that an ace and a 4 are drawn. –In symbols, find P(A 4) where A represents an ace of any suit, and 4 represents a four of any suit –Think what this question means, and solve. We are asking about a compound event. And there is no given. An ace and a 4 must be drawn to be a success, in either order. Essentially this gives us two cases: A or 4 1 A 2. This is a very compound event, two ands with an or between (or two intersections with a union between). Quite often, we handle compound events through the formulas weve seen, for union & intersection.

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7.5 Conditional Probability; Independent Events –#12) Two cards are drawn without replacement from an ordinary deck. Find the probability that an ace and a 4 are drawn. –In symbols, find P(A 4) where A represents an ace of any suit, and 4 represents a four of any suit P(A 4) = P( A A 2 ) = P( A ) + P( 4 1 A 2 ) – P( A A 2 ) = P( A ) + P( 4 1 A 2 ) = P( A 1 ) P( 4 2 |A 1 ) + P( 4 1 ) P( A 2 |4 1 ) = 4/52 4/51 + 4/52 4/51 = 16/ /2652 = 32/2652 = 8/ P(E F ) = P(E) + P(F) – P(E F ) mut. excl. P(E F)=P(E) P(F|E)

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7.5 Conditional Probability; Independent Events As said before, –Tree diagrams are handy for working with conditional probabilities and especially for illustrating such –And in particular, probability tree diagrams, with probabilities written on the branches, and then the path probabilities calculated and written for each of the paths, are extremely helpful –Now that weve developed the use of formulae further, lets look at the previous two problems done with a probability tree diagram to illustrate the idea. [on board]

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7.5 Conditional Probability; Independent Events –For the probability of the compound event of two events happening in succession, we have formulae such as P(A B ) = P(A) P(B|A) –But sometimes it happens that P(B|A) = P(B). In other words, the probability of Bs occurring is the same whether additional information about As having occurred or not is known or not In this situation, A and B are said to be independent events. –When events A and B are independent, the formula above reduces to P(A B ) = P(A) P(B)

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7.5 Conditional Probability; Independent Events –The formula P(A B ) = P(A) P(B) can be used when we know the two events are independent, but often this idea is taken in reverse, and –The formula is used to TEST whether A and B are independent –Each of the three probabilities is calculated separately (be especially careful with the first!) and then put together to see if the L.H.S. is equal to the R.H.S. for the equation P(A B ) = P(A) P(B).

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7.5 Conditional Probability; Independent Events P(A B ) = P(A) P(B) ? If P(A) =.5, P(B) =.6, P(A B) =.8, then A&B are independent … –P(A = P(A) + P(B) – P(A B) = =.3 –.3 = (.5) (.6) is true If P(A) =.5, P(B) =.6, P(A B) =.3, then A&B are not independent (they are dependent) … –P(A = P(A) - P(A B) = =.2 –.2 = (.5) (.6) is NOT true Use calculations similar to these, not guessing, to determine independence of events!

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7.5 Conditional Probability; Independent Events –Problem 7.5 # 46. The following table, based on data from the Centers for Disease Control, gives the number of new cases of the AIDS virus for men and women in the United States in 1998 by method of transmission. –Find (c) the probability that a newly diagnosed person with AIDS is a female. Method of transmn.MaleFemaleTotal Homosexual contact8,3880 Heterosexual contact1,1461,8062,952 Intravenous drug use3,6521,5415,193 Other5,2372,0547,291 Total18,4235,40123,824

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7.5 Conditional Probability; Independent Events –Problem 7.5 # 46 (c) the probability that a newly diagnosed person with AIDS is a female. –The table represents ALL the individual residents. –We are NOT told any additional information. So there is no given. We are simply asked about the gender of the patient. Method of transmn.MaleFemaleTotal Homosexual contact8,3880 Heterosexual contact1,1461,8062,952 Intravenous drug use3,6521,5415,193 Other5,2372,0547,291 Total18,4235,40123,824

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7.5 Conditional Probability; Independent Events –This time, not P(A|B) but only P(A) is asked! –P(Female) or –P(F) = n (F) / n(S) = 5401 / 23,824 Method of transmn.MaleFemaleTotal Homosexual contact8,3880 Heterosexual contact1,1461,8062,952 Intravenous drug use3,6521,5415,193 Other5,2372,0547,291 Total18,4235,40123, or 22.67%

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7.5 Conditional Probability; Independent Events –Problem 7.5 # 46. The following table, based on data from the Centers for Disease Control, gives the number of new cases of the AIDS virus for men and women in the United States in 1998 by method of transmission. –Find (d) the probability that a newly diagnosed person who contracted AIDS via heterosexual contact was a female. Method of transmn.MaleFemaleTotal Homosexual contact8,3880 Heterosexual contact1,1461,8062,952 Intravenous drug use3,6521,5415,193 Other5,2372,0547,291 Total18,4235,40123,824

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7.5 Conditional Probability; Independent Events –Problem 7.5 # 46 (d) the probability that a newly diagnosed person who contracted AIDS via heterosexual contact was a female. –The table represents ALL the individual residents. –We are told that the person contracted AIDS via heterosexual contact. This is additional information. Its the given. Then we are asked about the gender of the person. Method of transmn.MaleFemaleTotal Homosexual contact8,3880 Heterosexual contact1,1461,8062,952 Intravenous drug use3,6521,5415,193 Other5,2372,0547,291 Total18,4235,40123,824

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7.5 Conditional Probability; Independent Events –Our P(A|B) = n(A B) / n(B) becomes –P(intravenous drug use | Female) or –P(IV drug | F) = n ( IV drug F) / n(F) = 1541 / 5401 Method of transmn.MaleFemaleTotal Homosexual contact8,3880 Heterosexual contact1,1461,8062,952 Intravenous drug use3,6521,5415,193 Other5,2372,0547,291 Total18,4235,40123, or 28.53%

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