Let the number of white balls be n The number of red balls = 8 Now, the probability of drawing a white ball = \(\frac {n}{n + 8}\) The probability of drawing a red ball = \(\frac {8}{n + 8}\) Since Pr(White ball) = \(\frac{1}{2}\) x Pr(Red ball) \(\therefore \frac {n}{n + 8} = \frac {1}{2}\times \frac {8}{n + 8}\) = \(\frac {n}{n + 8} = \frac {4}{n + 8}\) \(\therefore n = 4\) Number of white balls = 4 The possible number of outcomes = n + 8 = 4 + 8 = 12 Pr(Red ball and White ball) = Pr(Red ball) x Pr(White ball) Pr(Red ball) = \(\frac {8}{12}\) Pr(Red ball) = \(\frac {4}{11}\) Pr(Red ball and white ball) = \(\frac {8}{12} \times \frac {4}{11}\) \(\therefore\)Pr(Red ball and white ball) = \(\frac {8}{33}\)