Question

A bag has __ green balls and 8 blue balls. If the probability of getting a blue ball is 1/3. How many green balls are in bag ? 16. 3 5. 8 ​

Answer 1

Step-by-step explanation:

A bag contains 6 red balls, 8 white balls, 5 green balls and 2 black balls. One ball is drawn at random from the bag.

Then total number of balls in bag =6+8+5+2+1=22

Total number of out come =22

(i)The event is getting a White ball=number of white ball in bag=8

Then probability= 8/22=4/11

(ii)The event getting the Red or black=The number of red & black ball in bag=6+8+1=9 (one ball is random)

The probability= 9/22

(iii)The event getting the Not green is total number of ball other then green ball=6+8+2+1=17

The probability= 17/22

(iv) The event getting the Neither white nor black=Total number of balls without white or black ball in bag=6+5=11

The probability= 11/22

= 1/2

Answer 2

Answer:

let green ball be x

p(blue ball) =1/3

p(green ball)= 1-1/3 = 2/3

no. of blue ball = 8

so,

total ball = x+8

then,

p(green)= x/x+8=2/3

3x=2(x+8)

3x-2x = 16

x=16

hence , no. of green balls be 16

Step-by-step explanation:

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