Probability

Let number of balls in first bag be x and second be y and in third be z First bag will contain three balls then, x=3 ,3+y+z=8 y+z=5 where y,z can take whole numbers Possible solutions are (0,5),(5,0),(1,4),(4,1),(2,3),(3,2) =6 Total possible ways of putting balls in bag are x+y+z=8 ,where x,y,z can take whole numbers On plotting different whole number values for above linear equation we get total 45 solutions  Probability that the first bag will contain 33 balls=6/45=2/15=0.133

Aakash Raavi

CMA Inter

★ 320

07-Jun-21 09:58

The solution suggested by Aakash is for identical balls. For different balls you to use selection/ combination approach.

Pavan Kumar

Faculty

09-Jun-21 00:04