In A Box, There Are 8 Red, 7 Blue And 6 Green Balls. One Ball Is Pick

1.	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?
Answer» 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) = n(E)/n(S) = 7/21 = 1/3. 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) = n(E)/n(S) = 7/21 = 1/3.

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?

Answer»

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) = n(E)/n(S) = 7/21 = 1/3.

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) = n(E)/n(S) = 7/21 = 1/3.