Discussion :: Probability
-
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?
|
A.
\( \frac { 1 } { 3 }\) |
|
B.
\( \frac { 3 } {4 } \) |
|
C.
\( \frac { 7 } {19 } \) |
|
D.
\(\frac {8 } { 21 } \) |
|
E.
\(\frac {9 } { 21 } \) |
Whatsapp
Facebook
Answer : Option A
Explanation :
Total number of balls = (8 + 7 + 6) = 21.
| Let E | = event that the ball drawn is neither red nor green |
| = event that the ball drawn is blue. |
n(E) = 7.
P(E) = \( \frac { n(E) } { n(S)}\)=\(\frac {7} { 21 } \) =\( \frac { 2 } { 3 }\)
Be The First To Comment