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