The probability of AL's number will be a whole number multiple of bill's and bill's number will be a whole number multiple of call's is 1÷ 80
The problem can be solved using Butte force approach
We can solves this in two cases
If Cal's number is :
If Bill's number is , Al’s can be any of 4, 6, 8, 10.
If Bill's number is , Al’s can be any of 6, 9.
If Bill's number is , Al’s can be 8.
If Bill's number is , Al’s can be 10.
Otherwise, Al's number could not be a whole number multiple of Bill's.
If Cal's number is 2 :
If Bill’s number is , Al's can be 8.
Otherwise, Al’s number could not be a whole number multiple of Bill's while Bill’s number is still a whole number multiple of Cal’s.
Otherwise, Bill's number must be greater than , i.e. Al’s number could not be a whole number multiple of Bill’s.
Clearly, there are exactly cases where Al's number will be a whole number multiple of Bill's and Bill's number will be a whole number multiple of Cal’s. Since there are 10×9×8 possible permutations of the numbers Al, Bill, and Cal were assigned, the probability that this is true is
9÷10×9×8
C = 1÷80
A similar problem is given at
https://brainly.com/question/15560501
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