First slide

Introduction to probability

Question

The probability that when 12 distinct balls are distributed among three boxes, the first will contain exactly three balls is

Moderate

A

$\frac{2^{9}}{3^{12}}$
B

$\frac{^{12} C_{3} \cdot 2^{9}}{9^{6}}$
C

$\frac{^{12} C_{3} \cdot 2^{12}}{3^{12}}$
D

$\frac{^{12} C_{3} \cdot 3^{9}}{3^{12}}$

Solution

3 balls for 1^st Box can be selected in $^{12} C_{3}$ ways, remaining 9 balls can be placed in remaining 2 boxes in 2⁹ ways.
Number of favourable cases $=^{12} C_{3} \times 2^{9}$
Total number of cases = 3¹²
Required Probability $= \frac{^{12} C_{3} \times 2^{9}}{3^{12}} = \frac{^{12} C_{3} \times 2^{9}}{9^{6}}$

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App