A bag contains 3 red, 3 white and 3 green balls. One ball is taken out of the bag at random. What is the probability that the ball drawn is

Answer

A, B, C

Explanation

The sample space
`S={R_(1),R_(2),R_(3),W_(1),W_(2),W_(3),G_(1),G_(2),G_(3)}`
`thereforen(S)=9.`
A is the event that the ball picked up is a red ball. Now there are three red balls out of 9 balls.
`thereforeA={R_(1),R_(2),R_(3)}`
`thereforen(A)=3.`
`P(A)=(n(A))/(n(S))=(3)/(9)=(1)/(3).`
B is the event that the ball picked up is not a red ball,i.e.a ball picked up is either white or green.
There are 3+3 =6 balls (white+green ) out of the 9 balls.
`thereforeB={W_(1),W_(2),W_(3),G_(1),G_(2),G_(3)}`
`thereforen(B)=6.`
`P(B)=(n(B))/(n(S))=(6)/(9)=(2)/(3)`
C is the event that the ball picked up is either red or white.
There are 3+3=6 balls which are either red or white.
`thereforeC={R_(1),R_(2),R_(3),W_(1),W_(2),W_(3)}`
`thereforen(C)=6.`
`P(C)=(n(C))/(n(S))=(6)/(9)=(2)/(3).`