Let X = {1,2,3,4} and Y = {1,2,3,4,5,6,7} a) What is the probability that a func

Question

Let X = {1,2,3,4} and Y = {1,2,3,4,5,6,7}
a) What is the probability that a function f:X-->Y isone-to-one?

b) What is the probability that a function f:X-->Y containsonly even numbers in its image?

c) What is the probability that a relation from X to Y is afunction from X to Y also?
a) What is the probability that a function f:X-->Y isone-to-one?

b) What is the probability that a function f:X-->Y containsonly even numbers in its image?

c) What is the probability that a relation from X to Y is afunction from X to Y also?

Explanation / Answer

Total number of posible functions from X -> Y = Nf=7x7x7x7 [as 1,2,3,4 in X can map to any of the 7 elements inY] (a) For 1-1 function, if 1 in X maps to some number in Y (7choices) then 2,3,4 in X can't map to that number in Y. Similarlyif now 2 in X maps to some number in Y (other than the one beingmapped with 1, so 6 choices) 3,4 in X can't map to that number andso on. Total number of possible 1-1 functions from X -> Y =If = 7x6x5x4 Probablity of a function being injective or 1-1 =Nf/If = (7x6x5x4)/(7x7x7x7) = 120/343 (b) Even numbers in Y = Ye = {2, 4, 6} Total number of functions possible from X -> Ye =3x3x3x3 Therefore required probability = (3x3x3x3)/(7x7x7x7) = 81/2401 (c) A relation from X to Y is any subset of X x Y (cartesianproduct) number of elements in X x Y = |X| x |Y| = 4 x 7 = 28 Therefore total number of possible relations from X -> Y =Rf = 228 (number of subsets of a set with 28elements) Every function is a relation. Therefore, probability of a relation being a function =Nf / Rf = 7x7x7x7/228 =2401/228

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