Have been trying for the last three hours, and going nuts. Please provide HINTS only, not the solution (or answer).
I'm doing this by a dumb approach, way too much of calculations and excel madness. I'm looking for an intelligent shortcut (or many of them). Looking to cut down on things algebraically. Here's the question.
(Dice = plural of die, the thing with the numbers 1 to 6 on its surface.)
Q. 1000 fair and 1000 unfair dice are available in a bag. On a certain day, from 8 a.m. to 6 p.m., 200 dice are randomly selected from the bag every hour and rolled. The faces are recorded and summed, and the rolled dice are then thrown away, ensuring no replacement. The fair dice behave normally ALL the time, but the unfair dice follow a certain pattern:
(a) Between odd and even hours (e.g. between 9 am and 10 am), the unfair dice change their [1, 2, 3] faces to [4, 5, 6] faces for the whole hour. So they become a dice with [4, 5, 6, 4, 5, 6] as the faces.
(b) Between even and odd hours (e.g. between 8 am and 9 am), the unfair dice change their [4, 5, 6] faces to [1, 2, 3] faces for the whole hour. So they become a dice with [1, 2, 3, 1, 2, 3] as the faces.
(c) However, there is an overriding rule: If any of the two bounding hours is prime, then all other unfair dice patterns are discarded, and the unfair dice change ALL their faces to that prime number only. (e.g. [5, 5, 5, 5, 5, 5] from 4 pm to 5 pm and 5 pm to 6 pm.) If both bounding numbers are prime, then the higher one is shown.
Find the probability (correct to 5 decimal places) that the total sum of all throws from 8 am to 6 pm is (a) Even (b) Is a prime
(12 hour format to be followed after 12 noon. 1 o'clock is 1 o'clock, not 13:00 hours).