A box contains 14 white balls and 6 black balls. Find the probability of first drawing a black ball and then a white ball without replacement.
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Correct Answer: Option B
Explanation:
The probability of picking a black ball = \(\frac{6}{6 + 14} = \frac{6}{20}\)
Probability of a white ball without replacement = \(\frac{14}{19}\)
Probability of a black ball and then white without replacement = \(\frac{6}{20} \times \frac{14}{19} = \frac{21}{95} = 0.22\)
The probability of picking a black ball = \(\frac{6}{6 + 14} = \frac{6}{20}\)
Probability of a white ball without replacement = \(\frac{14}{19}\)
Probability of a black ball and then white without replacement = \(\frac{6}{20} \times \frac{14}{19} = \frac{21}{95} = 0.22\)