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A bag contains $30$ balls out of which '$m$' number of balls are blue in colour.
(i) Find the probability that a ball drawn at random from the bag is not blue.
(ii) If $6$ more blue balls are added in the bag, then the probability of drawing a blue ball will be $\frac{5}{4}$ times the probability of drawing a blue ball in the first case. Find the value of $m$.
(i) Find the probability that a ball drawn at random from the bag is not blue.
(ii) If $6$ more blue balls are added in the bag, then the probability of drawing a blue ball will be $\frac{5}{4}$ times the probability of drawing a blue ball in the first case. Find the value of $m$.
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Sol. (i) $P(\text{ball drawn is not blue}) = \frac{30 - m}{30}$ or $1 - \frac{m}{30}$ (I) (1 Mark)
(ii) Total number of balls now $= 36$
Number of blue balls now $= m + 6$
$P(\text{ball drawn is blue}) = \frac{m+6}{36}$ (II) (1 Mark)
According to question, $\frac{m+6}{36} = \frac{5}{4} \times \frac{m}{30}$ (III) ($\frac{1}{2}$ Mark)
$\Rightarrow m = 12$ (IV) ($\frac{1}{2}$ Mark)
(ii) Total number of balls now $= 36$
Number of blue balls now $= m + 6$
$P(\text{ball drawn is blue}) = \frac{m+6}{36}$ (II) (1 Mark)
According to question, $\frac{m+6}{36} = \frac{5}{4} \times \frac{m}{30}$ (III) ($\frac{1}{2}$ Mark)
$\Rightarrow m = 12$ (IV) ($\frac{1}{2}$ Mark)