Question

A box contains 12 balls out of which x are black.
1. If one ball is drawn at random then what is the probability that it will be a black ball?
2. If 6 more balls are added, then probability of black ball are double than that earlier.
Find x.
(Note: This is a confirmation question. I got the answer already. Just want to confirm the answer)

Answer 1

Solution :-

1. Let the total number of black balls initially be x

Total number of balls in the box = 12

P(getting a black ball) = x/12

2. Now, 6 black balls are put in the box, then the total number of balls

= 12 + 6 

= 18 balls

Then total number of black balls = (x + 6)

Now,

P(getting a black ball) = (x + 6)/18

According to the question.

2(x/12) = (x + 6)/18

2x/12 = (x + 6)/18

Cross multiplying

⇒ 36x = 12x + 72

⇒ 36x - 12x = 72

⇒ 24x = 72

x = 3

So, initially the number of black balls in the box was 3.

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Answer 2

1) x / 12

2) x / 6 = x + 6/ 18

=> 18x = 6x + x^2

=> 3x = x^2

=> x = 3