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fBYj[ Ç p\R" ¨µáê
fBYj[ Ç p\R" ¨µáê-(c) If Y =Xthen B∩Y =B∩X=Bso that ˇis just the identity function In this case, ˇis certainly a bijection Now suppose that Y≠X Then there exists some x∈Xsuch that x∉Y Then we have ˇ(g) =g=ˇ({x}), so ˇfails to be injective However, ˇis surjective because for any C∈P(Y) we have ˇ =C∩Y=CIndividuals applying for licensure for professions that require !



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# / # 0 ) 1 2 3 4 556 & & # 7 !!For random variables X and R defined in Example 25, find PX(x) and PR(r) In addition, find the following probabilities (a) PX = 0 (b) PX < 3 (c) PR > 1 Problem 222 Solution From Example 25, we can write the PMF of X and the PMF of R as PX (x) = 1/8 x = 0 3/8 x = 1 3/8 x = 2 1/8 x = 3 0 otherwise PR (r) = 1/4 r = 0 3/4 r = 2 0 Both f (1) and f (2)=2 There is c in (0,2) such that f' (c)=f (b)f (a)/ba By the MVT, there is a number c in (0, 1) such that f' (c) = f (1) f (0)/ (1 0) = 2/1 = 2 You can use the same idea to show that for another number c in (1, 2), f' (c) = 0 If you knew that f' was continuous, you could use the Intermediate Value Theorem to show
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