First slide

Introduction to probability

Question

Twelve balls are placed in three boxes. The probability that the first box contains three balls is

Moderate

A

$\frac{110}{9} {(\frac{2}{3})}^{10}$
B

$\frac{9}{110} {(\frac{2}{3})}^{10}$
C

$\frac{^{12} C_{3}}{12^{3}} \times 2^{9}$
D

$\frac{^{12} C_{3}}{3^{12}}$

Solution

Since each ball can be placed in any one of the 3 boxes, therefore there are 3 ways in which a ball can be placed in any one of the three boxes. Thus, there are 3¹² ways in which 12 balls can be placed in 3 boxes. The number of ways in which 3 balls out of 12 can be put in the box is $^{12} C_{3}$ . The remaining 9 balls can be placed in 2 boxes in 2⁹ ways. So, required probability is
$\frac{^{12} C_{3}}{3^{12}} 2^{9} = \frac{110}{9} {(\frac{2}{3})}^{10}$

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