A bag contains 8 red balls and some white balls... - JAMB Mathematics 2023 Question

Let's denote the total number of balls as \( T \), the number of red balls as \( R \), and the number of white balls as \( W \).

Given that the probability of drawing a white ball is half of the probability of drawing a red ball, we can set up the following equations:

\[ P(\text{White}) = \frac{1}{2} \times P(\text{Red}) \]

Since \( P(\text{White}) = \frac{W}{T} \) and \( P(\text{Red}) = \frac{R}{T} \), we have:

\[ \frac{W}{T} = \frac{1}{2} \times \frac{R}{T} \]

Now, we know that \( R = 8 \) (the number of red balls). We can substitute this value into the equation:

\[ \frac{W}{T} = \frac{1}{2} \times \frac{8}{T} \]

To find \( W \), we can solve for \( T \):

\[ W = \frac{1}{2} \times 8 = 4 \]

So, there are 4 white balls.

Now, for the probability of drawing a red ball and then a white ball without replacement:

\[ P(\text{Red and White}) = \frac{R}{T} \times \frac{W}{T-1} \]

Substituting the values:

\[ P(\text{Red and White}) = \frac{8}{T} \times \frac{4}{T-1} \]

Since \( T = 8 + 4 = 12 \), we can substitute this value:

\[ P(\text{Red and White}) = \frac{8}{12} \times \frac{4}{11} \]

Simplifying:

\[ P(\text{Red and White}) = \frac{2}{3} \times \frac{4}{11} \]

Therefore, the correct answer is \(\frac{8}{33}\)