There are 8 red, 7 blue, and 6 green balls in a box. A single ball is chosen at random. How likely is it that it is neither red nor green?
A. 1/3
B. 3/4
C.7/19
D. 8/21
Answer:A
There are 8 red, 7 blue, and 6 green balls in a box. A single ball is chosen at random. How likely is it that it is neither red nor green?
A. 1/3 B. 3/4 C.7/19 D. 8/21
Explanation: The total number of balls is 21 (8 + 7 + 6).Let E denote the possibility that the drawn ball is neither red nor green, and B denote the possibility that the drawn ball is blue.n(E) = n(E) = n(E) = P(E) = n(E)/n(S) = 7/21 = ⅓