A bag contains 5 red and 3 white balls. Three balls are drawn at random, one by one, without replacement. The probability that the first two balls are red and the third ball is white is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(\displaystyle\frac{5}{28}\)
Let \(R_1\), \(R_2\) be the events that the first and second balls are red and \(W_3\) the event that the third ball is white. By the multiplication theorem for three events:
\(P(R_1\cap R_2\cap W_3) = P(R_1)\,P(R_2|R_1)\,P(W_3|R_1\cap R_2)\)
Each draw leaves one ball fewer in the bag:
\( \displaystyle P(R_1) = \frac{5}{8},\quad P(R_2|R_1) = \frac{4}{7},\quad P(W_3|R_1\cap R_2) = \frac{3}{6} \)
So
\( \displaystyle P = \frac{5}{8}\times\frac{4}{7}\times\frac{3}{6} = \frac{60}{336} \)
\( \displaystyle = \frac{5}{28} \)
Why the other options are incorrect
Option (1) \(\displaystyle\frac{15}{28}\) — Incorrect. This is the probability of 2 red and 1 white in any order, \(3\times\frac{5}{28}\).
Option (2) \(\displaystyle\frac{75}{512}\) — Incorrect. This is \(\frac{5}{8}\times\frac{5}{8}\times\frac{3}{8}\), the answer with replacement.
Option (3) \(\displaystyle\frac{5}{14}\) — Incorrect. This is \(\frac{5}{8}\times\frac{4}{7}\), only the first two draws; the third draw is left out.