Question

A bag contains 12 white and 18 red balls. Two balls are drawn in succession without replacement. The probability that the first is red and second is white is:

• A. \(\frac{63}{145}\)
• B. \(\frac{36}{154}\)
• C. \(\frac{36}{144}\)
• D. \(\frac{36}{145}\)

Answer 1

The Correct Option is D

The problem involves determining the probability of drawing two balls in succession from a bag containing 12 white and 18 red balls, with the condition that the first is red and the second is white.

To solve this, calculate the probability of each event and multiply them due to the sequence of independent events:

1. Calculate the probability of drawing a red ball first:

\[ P(\text{Red first}) = \frac{\text{Number of red balls}}{\text{Total number of balls}} = \frac{18}{30} = \frac{3}{5} \]

2. After drawing a red ball, calculate the probability of drawing a white ball next:
3. The total number of balls is now 29, and the number of white balls remains 12, so:

\[ P(\text{White second | Red first}) = \frac{12}{29} \]

Thus, the combined probability of the sequence is computed by multiplying these probabilities:

\[ P(\text{Red first and White second}) = \frac{3}{5} \times \frac{12}{29} = \frac{36}{145} \]

This steps prove that the probability the first ball is red and the second ball is white is \(\frac{36}{145}\).