Question

a bag contains a bag contains 4 green 5 black and 3 blue ball and then probability of getting that one drawn ball is not Green is

Answer 1

Heyaa frnd☺
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Given,

Green balls =4
Black balls. =5
Blue balls. =3

=》The total number of balls = 12

W.K.T,

The probability of getting that one drawn ball is not green i.e,

=》Total balls - Green balls
= 12-4
= 8.

=>Probability= 8/12

=2/3.
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Hope you understand...✌✌

Answer 2

n(s)= (G1,G2,G3,G4,B1,B2,B3,B4,B5,BL1,BL2,BL3)

     n(S)=12

probabilaty that ball drawn is not green

Let A be an event of not green ball

therefore,

n(A)=(B1,B2,B3,B4,B5,BL1,BL2,BL3)

n(A)=8

therefore

P(A)=  n(A)/n(S)

         8/12

=        4/6

=       2/3

=        0.6666666

NOTE : B=black,G=green,BL=blue