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.
0.21
0.22
0.30
0.70
Correct answer is B
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 = \(\frac{6}{20} \times \frac{14}{19} = \frac{21}{95} = 0.22\)