Number of red balls in the bag = 5 As the probability of blue balls is double the probability of red balls, we have that number of blue balls is double the number of red balls.
∴ Blue balls = 5 × 2 = 10.
[!! Let the number of blue balls = x
Number of red balls = 5
Total no. of balls = x + 5
Total outcomes in drawing a ball at random = x + 5
Number of outcomes favourable to red ball = 5
∴ P(R) = \(\frac{5}{x+5}\)
from the problem,
P(B) = 2 × \(\frac{5}{x+5}\) = \(\frac{10}{x+5}\)
Also \(\frac{5}{x+5}\) + \(\frac{10}{x+5}\) = 1
[∵ P(E) + P(\(\overline{E}\)) = 1]
⇒ \(\frac{5+10}{x+5}\) = 1
⇒ \(\frac{15}{x+5}\) = 1
⇒ x + 5 = 15 ]
