There are 9! arrangements of nine distinguishable objects in a How many are there? We start as we have done before, by assuming all nine objectsДЄre distinguishable. Possible sequences which satisfy the requirement that we have 4 AвЂ™s, 2 T вЂ™s, 2 GвЂ™s, and 1Д¬. (c) Each specific sequence has the probability given above, but in this case there are many The probability of this specific sequence is (1/4)9 = 3.8 Г— 10в€’6. The probability of an A in position 1 is 1/4, ofДЄ in position 2 is 1/4, of A in position 3 is 1/4, of T in position 4 is 1/4, and so on. (a) Each base occurs with probability 1/4. (c) What is the probability of finding any sequence that has four AвЂ™s, two TвЂ™s, two GвЂ™s, (b) What is the probability of finding the sequence AAAAAAAAA through random (a) What is the probability of finding the sequence AAATCGAGT through random The same answer as above (but notice how much more tedious it is).ДЄssume that the four bases A, C, T, and G occur with equal likelihood in a DNA sequence Up all the appropriate possible mutually exclusive events.

Note that we could have solved part (a) the same way it would have required adding Is simple to read off the answer in this problem: we want aH and aDSM, but notice weДonвЂ™t care about UCSF. Represents independent events of, for example, aH and aDSM and aUCSF. With respect to each other therefore they are all added. Each of these events is mutually exclusive Where the first term is the probability of acceptance at all 3, the second term representsДЄcceptance at H and DSM but rejection at UCSF, the third term represents acceptanceДЄt H and UCSF but rejection at DSM, etc. = p(aH)p(aDSM)p(aUCSF) + p(aH)p(aDSM)p(rUCSF)

Therefore all possibleД¬ircumstances are taken into account by adding the mutually exclusive events together, Rejection and acceptance at H are mutually exclusive. (b) The simple answer is that this is the intersection of two independent events:ДЄ more mechanical approach to either part (a) or this part is to write out all the Therefore the probability of at least one acceptance Rejection at UCSF is p(rUCSF) = 1 в€’ 0.1 = 0.9. The probability of rejection at H is p(rH) = 1 в€’ 0.5 = 0.5. These events are independent, so we have the answer. P (r) = the probability that youвЂ™re rejectedДЄt H and at DSM and at UCSF. Probability can be put in the above terms. (YouвЂ™re either accepted somewhere or youвЂ™re not.) But this

Of acceptance somewhere, P (a), is P (a) = 1 в€’ P (r), where P (r) is the probability that Means that if one happens, the other cannot happen). This doesnвЂ™t help because these events are not mutually exclusive (mutually exclusive We want the probability of getting into H or DSM or UCSF. Therefore we try to structure the problem into an andДЄnd or problem. When events are mutually exclusive and you want the probability of events A or B, Independent, and you want the probability of events A and B, you can multiply them. (a) The simplest way to solve this problem is to recall that when probabilities are (b) What is the probability that you will be accepted by both Harvard and Duluth? (a) What is the probability that you get in somewhere (at least one acceptance)? You guess that the probabilities youвЂ™llД«e accepted are: p(UCSF) = 0.10, p(DSM) = 0.30, and p(H) = 0.50. You have applied to three medical schools: University of California at San Francisco (UCSF),Дuluth School of Mines (DSM), and Harvard (H).

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