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) = \(\frac {8}{33}\)