Question

One bag contains 6 red, 2 blue, and 3 yellow balls. A second bag contains 2 red, 4 blue, and 5 yellow balls. A third bag contains 3 red, 7 blue, and 1 yellow ball. One bag is selected at random. If 1 ball is drawn from the selected bag, what is the probability that the ball drawn is yellow? 

Answer 1

$Question:- \\One\space\space bag\space\space contains\space\space 6\space\space red,\space 2\space blue,\space and\space 3\space yellow\space balls.\space \\ A\space second\space bag\space contains\space 2\space red,\space 4\space blue,\space and\space 5\space yellow\space balls.\space \\ A\space third\space bag\space contains\space 3\space red,\space 7\space blue,\space and\space 1\space yellow\space ball.\space \\ One\space bag\space is\space selected\space at\space random.\\ \space If\space 1\space ball\space is\space drawn\space from \space the\space selected\space bag,\space what\\ \space is\space the\space probability\space that\space the \space ball\space drawn\space is\space yellow?\space \\ \\\space ------------------------------------\space\\ solution\\ given\\ I\space bag\space contains\space =6\space red,\space 2\space blue,\space and\space 3\space yellow\space balls\\ II\space bag\space contains\space =2\space red,\space 4\space blue,\space and\space 5\space yellow\space balls\\ III\space bag\space contains\space =3\space red,\space 7\space blue,\space and\space 1\space yellow\space balls\\ Now,\space One\space bag\space is\space selected\space at\space random\space and\space 1\space ball\space is\space drawn\space from\space the\space selected\space bag\\ Pr(the\space ball\space drawn\space is\space yellow)\\= Pr(I\space bag\space select).Pr(the\space ball\space drawn\space is\space yellow\space from\space I\space bag)+Pr(II\space bag\space select).Pr(the\space ball\space drawn\space is\space yellow\space from\space II\space bag)+Pr(III\space bag\space select).Pr(the\space ball\space drawn\space is\space yellow\space from\space III\space bag)\\ =\frac{1}{3}.\frac{3}{11}+\frac{1}{3}.\frac{5}{11}+\frac{1}{3}.\frac{1}{11}\\ =9/33\\ answer$