If x and y are inclusive ranges, can someone explain why is not possible if I take total cases of (x,y) as (0,0) (0,1).......
This will form 21 cases. Out of which, x>y in 3 cases (1,0), (2,0), (2,1). So the probability comes out to 1/7.
Please help me understand.
Kinshook wrote:
Let the divide the time line between 0 & 6 in 2 parts
Part 1: 0 to 2
Probability that x>y = 1/2 since x & y are equally probable, half the time x > y
Part 2: 2 to 6
Probability that x>y = 0 since x has no value in the interval
Since Part 1 (0 to 2) is 2/6th part of the complete time line of 0 to 6 and Part 2 (4 to 6) is 4/6th part of the complete timeline of 0 to 6,
Probability that x> y = (2/6)*(1/2) + (4/6)*0 = 1/6
sherxon wrote:
Kinshook wrote:
Asked: If a real number x is randomly chosen between 0 and 2, inclusive, and a real number y is randomly chosen between 0 and 6, inclusive, what is the probability that x>y?
For y > 2; x > y is not possible
For 0<=y<=2; It is 1/2 probability that x>y
The probability that x>y = (4/6)*0 + (2/6)*(1/2) = 1/6
IMO C
Can you elaborate on the calculation. Why are you taking 4/6 and 2/6? And why the probability is 1/2 for interval [0;2]?
I would be really grateful if you could answer.
Probability is my weakest area((