A bus arrives every 30 minutes at a specified stop beginning at 6:30 A. M. and continuing until 9:00 A. M. A certain passenger does not know the schedule but arrives randomly (uniformly distributed) between 7:00A. M. and 7:45A. M. every morning. What is the probability that the passenger waits A than X minutes for a bus? Also find the Expectation and Variance? Last Digit of Roll No. 0 1 2 3 4 5 6 7 8 9
The Value of A more more more more more Less Less Less Less Less
The Value of X 3 2 1 4 5 2 1 3 4 5

Ok here we go ..there are plenty of ways to solve such math problems but we will go for the smartest and easiest ..5x+2y=7  -2x+6y=9we rearrange them into the following ..by multiplying the 1st equation by 2 it becomes 10x + 4y = 14we rearrange it and so it becomes 10x = 14 - 4yas for the second equation we multiply it by 5 and rearrange it so it becomes as to the following ..-10x + 30y = 45rearranging so 10x = 30y -45now 10x equals both 14 - 4y and 30y -45so   14 - 4y = 30y -4531 = 26yy= 31/26 = 1.19 or around 1.2so by now we can calculate x easily as  -2x+6y=9so 2x = 6y -9 = 6 * 31/26 - 9 = -1.84  then x = - 0.92 or around -0.9best luck!