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KG NKG

Sixty percent of the tourists in London go to Buckingham Palace (BP) and forty percent visit Kensington Gardens (KG). What is the probability an individual will do both? a) 20% b) 24% c) 36% d) 100% This question is written in symbols as P(BP and KG).

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KG NKG

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  1. Sixty percent of the tourists in London go to Buckingham Palace (BP) and forty percent visit Kensington Gardens (KG). What is the probability an individual will do both? a) 20% b) 24% c) 36% d) 100% This question is written in symbols as P(BP and KG). In the context of a joint probability table we have KG NKG BP P(BP and KG) = ? .6 NBP .4 .4 .6 1 I originally said you solve by trial and error. Thus if P(BP and KG) = .2 then P(BP and NKG) = .4, P(NBP and KG) = .2, and P(NBP and NKG) = .2. This would give

  2. KG NKG BP .2 .4 .6 NBP .2 .2 .4 .4 .6 1 And everything seems right with the table. But, what if we had looked at answer b first in our trial and error. Thus if P(BP and KG) = .24 then P(BP and NKG) = .36, P(NBP and KG) = .16, and P(NBP and NKG) = .24. This would give KG NKG BP .24 .36 .6 NBP .16 .24 .4 .4 .6 1 And everything seems right with the table. But, what if we had looked at answer c first in our trial and error. Thus if P(BP and KG) = .36 then P(BP and NKG) = .24, P(NBP and KG) = .04, and P(NBP and NKG) = .36. This would give KG NKG BP .36 .24 .6 NBP .04 .16 .4 .4 .6 1 And everything seems right with the table.

  3. Here is where I went wrong. I assumed that all we had to look at during the trial and error was for the rows to add up to the row totals and the columns to add up to the column totals. This is necessary to have the right table and thus the right answer. But this is not sufficient (meaning not the only thing necessary!). Remember we are looking for the answer to the question P(BP and KG). Also note that P(BP and KG) = P(KG and BP), meaning on an intersection of events the order in which you write the events doesn’t matter. Now, in general, the definition of conditional probability is P(A│B) = P(A and B)/P(B), and P(B│A) = P(B and A)/P(A). But P(A and B) = P(B and A), so, P(B│A) = P(A and B)/P(A). In our specific problem we have P(BP│KG) = P(BP and KG)/P(KG), and P(KG│BP) = P(KG and BP)/P(BP). But P(BP and KG) = P(KG and BP), so, P(KG│BP) = P(BP and KG)/P(BP). So, from these definitions we have

  4. P(BP│KG) = P(BP and KG)/P(KG), and P(KG│BP) = P(BP and KG)/P(BP). In our question we want P(BP and KG). From the two definitions we have P(BP and KG) = P(KG)P(BP│KG) = .4P(BP│KG) and P(BP and KG) = P(BP)P(KG│BP) = .6P(KG│BP). This implies .4P(BP│KG) = .6P(KG│BP) and this is only true when P(BP│KG) = .6 and P(KG│BP) = .4. From the 3 scenarios on side 2 only the second one has these characteristics (Check them). Thus, the answer to the question is b) 24%.

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