Question

If a box has 8 red balls, 12 white balls, 17 black balls. If two balls are taken one by one without replacement, then the probability of taking one red ball and one black ball is

• A. \( \frac{136}{567} \)
• B. \( \frac{136}{561} \)
• C. \( \frac{80}{561} \)
• D. \( \frac{128}{561} \)

Hint:

When calculating probabilities with multiple events, use the multiplication rule and adjust for each step after removing or changing the total number of items.

Answer 1

The Correct Option is C

Given that the box contains: - 8 red balls - 12 white balls - 17 black balls The total number of balls in the box is \( 8 + 12 + 17 = 37 \). We are asked to find the probability of selecting one red ball and one black ball. The probability of selecting one red ball and one black ball can be found using the multiplication rule for probability. We first select one red ball, then one black ball. 1. The probability of selecting a red ball on the first draw is: \[ P(\text{red}) = \frac{8}{37} \] 2. After the red ball is removed, there are now 36 balls left, with 17 black balls remaining. So the probability of selecting a black ball on the second draw is: \[ P(\text{black}) = \frac{17}{36} \] 3. The total probability of selecting one red ball and one black ball is: \[ P(\text{red and black}) = \frac{8}{37} \times \frac{17}{36} = \frac{136}{1332} = \frac{80}{561} \] 
Thus, the correct answer is \( \frac{80}{561}  \).