Question

a box contains 12 balls from which X is black balls if one ball is drawn out what is the probability of getting black ball?.if 6more black balls are added the probability of black ball is now double than before find x​

Answer 1

$\large\underline{\sf{Solution-}}$

➢ Given that, in a bag

• Total number of balls in a bag = 12

• Number of black balls = x

We know,

$\sf \:Probability \: of \: an \: event =\dfrac{Number \: of \: favourable \: outcomes}{Total \: number \: of \: outcomes \: in \: sample \: space}$

So,

$\rm :\longmapsto\: \:P(getting \: black \: ball) =\dfrac{Number \: of \: black \: balls}{Total \: number \: of \: balls}$

$\rm :\longmapsto\: \:P(getting \: black \: ball) =\dfrac{x}{12} - - (1)$

Now,

➢ 6 more black balls are put in a bag.

So,

➢Total number of balls in a bag = 12 + 6 = 18

➢Number of black balls = x + 6

We know,

$\sf \:Probability \: of \: an \: event =\dfrac{Number \: of \: favourable \: outcomes}{Total \: number \: of \: outcomes \: in \: sample \: space}$

So,

$\rm :\longmapsto\: \:P(getting \: black \: ball) =\dfrac{Number \: of \: black \: balls}{Total \: number \: of \: balls}$

$\rm :\longmapsto\: \:P(getting \: black \: ball) =\dfrac{x + 6}{18} - - (2)$

According to statement,

Probability of getting black ball is double of getting black ball in first case.

$\rm :\longmapsto\:\dfrac{x + 6}{18} = 2 \times \dfrac{x}{12}$

$\rm :\longmapsto\:\dfrac{x + 6}{18} = \dfrac{x}{6}$

$\rm :\longmapsto\:\dfrac{x + 6}{3} = x$

$\rm :\longmapsto\:x + 6 = 3x$

$\rm :\longmapsto\: 6 = 3x - x$

$\rm :\longmapsto\: 6 = 2x$

$\bf\implies \:x = 3$

Explore more :-

• The sample space assocoated with any random experiment is the collection of all the possible outcomes.

• An event associated with a random experiment is a part of the sample space or may not be a part of sample space.

• The probability of any outcome is a number between 0 and 1 including 0 and 1.

• The probability of sure event is 1.

• The probability of impossible event is 0.