Question

A bag contains18 balls out of which x balls are red if one ball is drawn at random from the bag what is probability that it is not red and if 2 more red balls are put in the the probanility of drawinga red ball will be 9by 8 times the probability of drawing a red ball in first case fund the value of x

Answer 1

Total number of balls = 18

Number of red balls = $x$

Question #1 answer

Number of balls drawn = 1

Number of balls which are not red $=18-x$

∴ Probability of getting a non-red ball $=\dfrac{18-x}{18}$

Question #2 answer

Probability of drawing a red ball in the first case $=\dfrac{x}{18}$

Two more red balls are put in the bag.

So, total number of red balls $=x + 2$

So, total number of all balls $=18+2=20$

Probability of drawing a red ball in the second case = $\dfrac{x+2}{20}$

But it is also said that the second case probability = $\frac{9}{8} \times$ first case

⇒ $\dfrac{x+2}{20} = \dfrac{9}{8} \times \dfrac{x}{18}$

⇒ $\dfrac{x+2}{20} = \dfrac{9x}{8\times 18)}$

⇒ $\dfrac{x+2}{20} = \dfrac{9x}{144}$

We can cancel the denominators appropriately as follows:

⇒ $\dfrac{x+2}{5} = \dfrac{9x}{36}$

Now do cross multiplication of denominators:

⇒ $36 \left(x+2\right) = 5\times9x$

⇒ $36x+72 = 45x$

Now do transposition:

⇒ $36x-45x = -72$

⇒ $-9x = -72$

⇒ $x = \dfrac{-72}{-9}$

⇒ $x = 8$

So, value of x = 8.