0 0 votes Can someone explain answer for this question(26)? In a box, there are $8$ red, $7$ blue and $6$ green balls. One ball is picked up randomly. What is the probability that it is neither red nor green ? $\frac{2}{3}$ $\frac{3}{4}$ $\frac{7}{19}$ $\frac{9}{21}$ Quantitative Aptitude + – ranarajesh495 14 points 3.0k views answer comment Share Follow Print See 1 comment 1 1 comment reply Mk Utkarsh 256 points commented Dec 7, 2018 reply Follow flag all options are wrong answer is $\frac{1}{3}$ 1 1 replyShare Please log in or register to add a comment.
0 0 votes Total number of balls = 8 + 7 + 6 = 21 Let E be the event of selecting neither red nor green ball Then n(E) = 7 Required probability = 7/21 = 1/3 mudit23june answered Dec 8, 2018 mudit23june 348 points comment Share Follow 0 reply Please log in or register to add a comment.
