A number x ∈ R is chosen from the closed interval [ 2 , 1 0 ] , the probability that ⌊ 2 x ⌋ is even can be expressed as b a , where a and b are positive co-prime integers, find a + b
Details and Assumptions
⌊ x ⌋ is the floor function. ⌊ x ⌋ Is the largest integer not greater than x . For example: ⌊ 3 ⌋ = 3 , ⌊ 5 . 8 ⌋ = 5 , ⌊ π ⌋ = 3
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But wait. x is even when it belongs to [ 4 ; 6 ) (in, say, p cases) or [ 8 ; 1 0 ) (in p cases) and odd when it belongs to [ 2 ; 4 ) (in p cases) or [ 6 ; 8 ) (in p cases) or it is equal to 1 0 (in one case). Therefore, the probability that it is even is 2 p + 1 p , which is not exactly 2 1 . If the probability is indeed 2 1 , I'd say that mathematics is weird and doesn't make sense in this case...
hey "x" is a number not an integer so how can you take sum of lengths of interval =4
There are total 8 ranges... X/2 belongs to [2,3), [3,4)...upto [9,10].....and from these... Only 4 will give even value.... So probability will be 4/8 = 1/2 a+b = 3
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The length of the interval [ 2 , 1 0 ] is 8 units. If 4 ≤ x < 6 or 8 ≤ x < 1 0 then ⌊ 2 x ⌋ is an even integer. The sum of the lengths of the interval [ 4 , 6 ) and [ 8 , 1 0 ) is 4.
By Geometric Probability, the probability that ⌊ 2 x ⌋ is even is 8 4 = 2 1 . Therefore a = 1 and b = 2 , and a + b = 3