0 votes 0 votes Suppose that a solution $(X,Y,Z)$ of the equation $X+Y+Z=20,$with $X,Y$ and $Z$ non-negative integers, is chosen at random.What is the probability that $X$ is divisible by $5?$ $A)\frac{1}{4}$ $B)\frac{5}{21}$ $C)\frac{2}{7}$ $D)\frac{3}{20}$ Combinatory discrete-mathematics combinatory + – Lakshman Bhaiya asked Oct 24, 2018 Lakshman Bhaiya 390 views answer comment Share Follow See all 5 Comments See all 5 5 Comments reply Show 2 previous comments Lakshman Bhaiya commented Oct 24, 2018 reply Follow Share yes can you explain the procedure? 0 votes 0 votes akash.dinkar12 commented Oct 24, 2018 reply Follow Share the procedure will be in this way: Number of non-negative integral solutions = 22C2 For finding X for divisible by 5: X = 0 then Y + Z = 20 Number of non negative integral solutions = 21C1 = 21 X =5 then Y+ Z = 15 Number of non negative integral solutions = 16C1 = 16 X=10 then Y + Z = 10 Number of non negative integral solutions =11C1 = 11 X =15 then Y + Z = 5 Number of non negative integral solutions = 6C1 = 6 X =20 then Y + Z =0 Number of non negative integral solutions = 1C1 = 1 Sum of all these values = 21 + 16 + 11 + 6 + 1 = 55 Required probability = 55 / 22C2 = 5/21 4 votes 4 votes Lakshman Bhaiya commented Oct 24, 2018 reply Follow Share Thanks 0 votes 0 votes Please log in or register to add a comment.