Question

A bag contains 18 balls out of which x balls are red.

(i) If one ball is drawn at random from the bag, what is the probability that it is not red?
(ii) If 2 more red balls are put in the bag, the probability of drawing a red ball will be `9/8` times the probability of drawing a red ball in the first case. Find the value of x.

Answer 1

(i)
The total number of balls in the bag is 18.
∴ Total number of outcomes = 18
Let E be the event of drawing a non-red ball from the bag.
There are x red balls in the bag. Therefore, there are (18 − x) non-red balls in the bag.
So, number of favourable outcomes = 18 − x

∴ P(E) =\[\frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}\]=\[\frac{18 - x}{18}\]

(ii)
Let F be the events of drawing a red from the bag in the first case and G be the event of drawing a red from the bag in this case.

P(F) =\[\frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}\]=\[\frac{x}{18}\]

When 2 more red balls are put in the bag, then the total number of balls in the bag is 20.
∴ Total number of outcomes = 20
There are (x + 2) red balls in the bag.
So, number of favourable outcomes = x + 2

P(G)=\[\frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}\]=\[\frac{x + 2}{20}\]

It is given that

P(G) =\[\frac{9}{8}\]98P(F)

\[\therefore \frac{x + 2}{20} = \frac{9}{8} \times \frac{x}{18}\]
\[ \Rightarrow \frac{x + 2}{20} = \frac{x}{16}\]
\[ \Rightarrow 16x + 32 = 20x\]
\[ \Rightarrow 4x = 32\]
\[ \Rightarrow x = 8\]

Hence, the value of x is 8.