For what fraction of a day wil a 12 hour digital clock that shows only the hours and minutes display at least one 5 or 6
+6248 from 5:00 - 6:59 there will be a 5 or 6 visible at all times
from H:50 - H:59 there will be a 5 visible at all times, 1<= H <= 4 and 7 <= H <= 12
from H:M5 - H:M7 there will be a 5 or 6 visible at all times, 1 <= H <= 4 and 7<= H <= 12, 0 <= M <= 4
Summing all this up
2 hrs = 120 minutes for the first category
10 minutes apiece of 10 hrs = 100 min for the second category
2 minutes 5 times an hr for 10 hrs = 100 minutes
So a total of 320 minutes out of a total of 12 x 60 = 720 minutes
p = 320/720 = 4/9
+6248 from 5:00 - 6:59 there will be a 5 or 6 visible at all times
from H:50 - H:59 there will be a 5 visible at all times, 1<= H <= 4 and 7 <= H <= 12
from H:M5 - H:M7 there will be a 5 or 6 visible at all times, 1 <= H <= 4 and 7<= H <= 12, 0 <= M <= 4
Summing all this up
2 hrs = 120 minutes for the first category
10 minutes apiece of 10 hrs = 100 min for the second category
2 minutes 5 times an hr for 10 hrs = 100 minutes
So a total of 320 minutes out of a total of 12 x 60 = 720 minutes
p = 320/720 = 4/9
