Twelve balls are distributed among three axes. The probability that the first box contains 3 balls is:
Answer & explanation
Correct answer: option 1
Total arrangement = 312 [As each ball have '3' choices]
Favorable ways = (Choose '3' balls out of 12 and place in Ist box) × (Remaining balls each has 2 choices)
$={^{12}C}_3×2^9$ ⇒ P(required) = $\frac{{^{12}C}_3×2^9}{3^{12}}=\frac{110×2^{10}}{3^{12}}$