A box contains 8 pens, of which 3 are blue and 5 are black. Two pens are taken out at random, one after the other, without replacement. The probability that the first pen is blue and the second pen is black is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(\displaystyle\frac{15}{56}\)
Let E be the event that the first pen is blue and F the event that the second pen is black. By the multiplication theorem, \(P(E\cap F) = P(E)\,P(F|E)\).
\( \displaystyle P(E) = \frac{3}{8} \)
After a blue pen is taken out, 7 pens are left, of which 5 are black:
\( \displaystyle P(F|E) = \frac{5}{7} \)
So
\( \displaystyle P(E\cap F) = \frac{3}{8}\times\frac{5}{7} = \frac{15}{56} \)
Why the other options are incorrect
Option (2) \(\displaystyle\frac{15}{64}\) — Incorrect. This is \(\frac{3}{8}\times\frac{5}{8}\), the answer with replacement.
Option (3) \(\displaystyle\frac{15}{28}\) — Incorrect. This is the probability of one blue and one black pen in either order, \(2\times\frac{15}{56}\).
Option (4) \(\displaystyle\frac{3}{28}\) — Incorrect. This is the probability that both pens are blue, \(\frac{3}{8}\times\frac{2}{7}\).