A box contain 5 white and 8 black balls. Two successive drawings of three balls at a time are made such that the balls are not replaced before the second draw. The probability that the first draw gives 3 white balls and second draw gives 3 black balls is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( \displaystyle\frac{7}{429} \)
By the multiplication rule, the probability is P(first draw 3 white) × P(second draw 3 black, given the first).
First draw, from 13 balls:
\( \displaystyle P(3\text{ white})=\frac{^{5}C_{3}}{^{13}C_{3}} \)
\( \displaystyle =\frac{10}{286}=\frac{5}{143} \)
After this, 10 balls remain: 2 white and 8 black. Second draw:
\( \displaystyle P(3\text{ black})=\frac{^{8}C_{3}}{^{10}C_{3}} \)
\( \displaystyle =\frac{56}{120}=\frac{7}{15} \)
\( \displaystyle P=\frac{5}{143}\times\frac{7}{15}=\frac{7}{429} \)