Continued from above.
Because of the buckets of slop, it’s not easy to tell that Jojo’s answer is the same as Melody’s result; it’s still the wrong answer. The reason is the technique used for the solution does not represent the method Grogg uses to select which numbers are colored.
My answer is spot on. I can “prove” my answer is correct, though not with precise mathematics, because I lack the proficiency. (Lancelot could explain why in a blink.) My solution method follows the selection schemes (0-6) Grogg uses to color the numbers. Choose 0 through 6 numbers (for choices of 1-5 the numbers are selected randomly), color them, then test if it’s factorific. Each scheme has an individual probability of producing a factorific coloring.
Taking the sum of probabilities for each individual scheme and multiplying by a scheme’s selection probability (1/7), gives the true probability of selecting a factorific coloring. This is different from assuming a number has a 50% probability of being blue. The net change is 75% greater than the ratio of the number of success to the total sample space (1/3) vs, (1/4).
This question and others like it is interesting to me because a random selection process produces a success rate (factorific coloring). This question has minimal interest. There is no comment from the person who asked it. There are only 17 additional views since I posted my answer. Melody has not commented on it –she may not have seen it.
When Lancelot Link was teaching me combinatorics and statistics, he explained that there were many statistics questions where the “natural” solution was wrong. The most famous of these is the “Monty Hall” problem. The correct solution to this problem eluded many highly educated and experienced statisticians. Many disagreed with the correct solution after its presentation. Lancelot said though this is quantifiable by using Bayesian statistics, it was easy enough for any dumb dumb to construct a Monte Carlo simulation to demonstrate the correct solution.
This question does not require Bayesian statistics to solve –a simple weighting of the expectations solves this. To be sure that I’m not full of blarney and BS (which is becoming epidemic on here), I will write a Monte Carlo simulation. I’m certain the results will hover very close to the calculated (1/3) probability. Further, analyzing the data will show a mostly flat distribution of the numbers—in both the selected sets (0-6) and selected numbers (1-6).
GA
