Question

A bag contains $15$ balls of which $x$ are blue and the remaining are red . if the number of red balls are increased by $5$, the probability of drawing the red ball doubles find
(i) P(red ball)
(ii) P(blue ball)
(iii) p(blue ball if 5 balls are actually added).

Answer 1

$\Rightarrow$ Number of blue balls $=15$

$\Rightarrow$ Number of blue balss $=x$

$\Rightarrow$ Number of red balls $=15-x$

$\therefore$ Probability of taking a red ball $=\frac{15-x}{15}$

Now, after $5$ red balls are added,

$\Rightarrow$ Number of balls $=15+5=20$

$\Rightarrow$ Number of blue balls $=x$

$\Rightarrow$ Number of red balls $=(15-x)+5=20-x$

$\Rightarrow$ Probability of taking a red ball $=\frac{20-x}{20}$

According to the question,

$\frac{15-x}{15}\times 2=\frac{20-x}{20}$

$\frac{30-2x}{15}=\frac{20-x}{20}$

$600-40x=300-15x$

$300=25x$

$\therefore$ $x=\frac{300}{25}=12$

$\therefore$ We have $12$ blue balls and $3$ red balls .

$\left(i\right)$ $P\left(redball\right)=\frac{3}{15}=\frac{1}{5}$

$\left(ii\right)$ $P\left(blueballs\right)=\frac{12}{15}$

$\left(iii\right)$ $P\left(blueballif5ballsareactuallyadded\right)=\frac{12}{15+5}=\frac{12}{20}=\frac{3}{5}$