To find the probability of drawing a second white ball given that the first ball drawn was white, we will use conditional probability. This problem can be framed as finding [tex]\( P(B|A) \)[/tex], where:
- [tex]\( A \)[/tex] is the event of drawing a white ball first.
- [tex]\( B \)[/tex] is the event of drawing a white ball second.
We need to find [tex]\( P(B|A) \)[/tex], which is the conditional probability of [tex]\( B \)[/tex] given [tex]\( A \)[/tex]. According to the formula for conditional probability:
[tex]\[ P(B|A) = \frac{P(A \text{ and } B)}{P(A)} \][/tex]
We have the following information provided in the problem:
- [tex]\( P(A \text{ and } B) = \frac{1}{5} \)[/tex] (the probability of drawing two white balls consecutively).
- [tex]\( P(A) = \frac{7}{15} \)[/tex] (the probability of drawing a white ball first).
Now, substituting these values into the formula for conditional probability:
[tex]\[ P(B|A) = \frac{\frac{1}{5}}{\frac{7}{15}} \][/tex]
This simplifies to:
[tex]\[ P(B|A) = \frac{1/5}{7/15} = \frac{1/5 \times 15/7}{1} = \frac{15}{35} = \frac{3}{7} \][/tex]
Hence, the probability of drawing a second white ball given that the first ball drawn was white is [tex]\(\frac{3}{7}\)[/tex].
Therefore, the correct answer is [tex]\(\frac{3}{7}\)[/tex].
