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A question from RHE is: "What is the value of the following expression? Math.round(Math.random() + 2.50001);" Answers: A) 2; B) 3; C) It is impossible to say. The answer is 3 (not "It is impossible to say"). This implies that you cannot get a result of 3.5 The implication of this is that Math.random() won't return .99999. I realise it's result is 0<=x<1.0. So is the result only to four decimal places (ie max value is 0.9999)?

Originally posted by Trevor Green: A question from RHE is: "What is the value of the following expression? Math.round(Math.random() + 2.50001);" Answers: A) 2; B) 3; C) It is impossible to say. The answer is 3 (not "It is impossible to say"). This implies that you cannot get a result of 3.5 The implication of this is that Math.random() won't return .99999. I realise it's result is 0<=x<1.0. So is the result only to four decimal places (ie max value is 0.9999)?

Hi, Math.random() results in double primitive type and you get result to approx. 12 decimal places but it will always be less than 1 so after adding to 2.50001 it will always be very close to 3 but never equal to 3, and Math.round() results in long primitive type so you get 3 I hope its clear Jyotsna

Trevor Green
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If Math.random() returns a double to 12 decimal places, then 0.999999999999 is possible. 2.50001 + 0.99999999999 = 3.5000009999 This would round to 4. But the answer is B) 3. Isn't it c) It is impossible to say, because, in this rarest of cases, it could also be 4, as well as 3?