Question

a bag contain 6 yellow balls,3 red balls and 2 green balls,In how many ways can 5 balls drawn from the bag if at least one yellow ball is to be included in the draw?
a)464
b)463
c)462
d)461

Answer 1

Answer:

The answer is

462

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Answer 2

Answer:

Yellow balls =6

Red balls =3

Green balls =2

Total number of balls =11

5 balls to be drawn and atleast 1 is yellow

Therefore

the probability of drawing 5 yellow balls =6C5 =6*5*4*3*2/5*4*3*2*1

=6

The probability of drawing

(4y+1r) =6C4*3C1=15*3=45

(4y+1g) =6C4*2C1=15*2=30

(3y+2r) =6C3*3C2=20*3=60

(3y+1r+1g) =6C3*3C1*2C1=20*3*2=120

(3y+2g) =6C3*2C2=20*1=20

(2y+3r) =6C2*3C3=15*1=15

(2y+2r+1g) =6C2*3C2*2C1=15*3*2=90

(2y+1r+2g) =6C2*3C1*2C2=15*3*1=45

(1y+3r+1g) =6C1*3C3*2C1=6*1*2=12

(1y+2r+2g) =6C1*3C2*2C2=6*3*1=18

Therefore the total probability

=6+45+30+60+120+20+15+90+45+12+18= 461

D. 461 is the answer