Permutaions-five distinguishable balls

parveen110 wrote:Five distinguishable balls are placed into three distinguishable boxes. What is the probablity that atleast one box is empty?

P(at least 1 box is empty) = 1 - P(no box is empty).

Total ways to distribute the balls:
Number of options for the 1st ball = 3. (Any of the 3 boxes.)
Number of options for the 2nd ball = 3. (Any of the 3 boxes.)
Number of options for the 3rd ball = 3. (Any of the 3 boxes.)
Number of options for the 4th ball = 3. (Any of the 3 boxes.)
Number of options for the 5th ball = 3. (Any of the 3 boxes.)
To combine these options, we multiply:
3*3*3*3*3 = 3�.

There are 2 cases in which no box will be empty.

Case 1: One special box gets 3 balls, the other 2 boxes get 1 ball each
Number of box options for the special box = 3. (Any of the 3 boxes.)
From the 5 balls, the number of ways to choose 3 balls for the special box = 5C3 = (5*4*3)/(3*2*1) = 10.
Number of ball options for the next box = 2. (Either of the 2 remaining balls.)
Number of ball options for the last box = 1. (Only 1 ball left.)
To combine these options, we multiply:
3*10*2*1 = 60.

Case 2: One special box gets 1 ball, the other 2 boxes get 2 balls each
Number of box options for the special box = 3. (Any of the 3 boxes.)
Number of ball options for the special box = 5. (Any of the 5 balls.)
From the 4 remaining balls, the number of ways to choose 2 balls for the next box = 4C2 = (4*3)/(2*1) = 6.
From the 2 remaining balls, the number of ways to choose 2 balls for the last box = 2C2 = (2*3)/(2*1) = 1.
To combine these options, we multiply:
3*5*6*1 = 90.

Thus:
P(no box is empty) = (60+90)/3� = 50/81.
Thus:
P(at least 1 box is empty) = 1 - 50/81 = [spoiler]31/81[/spoiler].

Here's an easier but more treacherous way:

Suppose we want the probability that Box 1 is empty. When we place the first ball, the probability is (2/3) that goes in either Box 2 or Box 3. Since we want each of the five balls to go in Box 2 or Box 3, the probability is really (2/3)�.

Continuing like this, the probability that Box 2 is empty is (2/3)� and the probability that Box 3 is empty is (2/3)�.

However, we've overcounted! The probability that Boxes 1 and 2 are both empty appears TWICE (in the probability for Box 1 and the probability for Box 2). So we have to subtract that ONCE. The probability that Boxes 1 and 2 are both empty is (1/3)�, as all the balls would have to go in Box 3.

The probabilities for Box 2 and Box 3 both empty and Box 1 and Box 3 both empty are likewise (1/3)�.

The probability that all three boxes are empty is 0, as we have to pull the balls in SOME box.

Hence we have

Prob(1 empty) + Prob(2 empty) + Prob(3 empty) - Prob(1+2 empty) - Prob(1+3 empty) - Prob(2+3 empty) + Prob(all empty) =

(2/3)� + (2/3)� + (2/3)� - (1/3)� - (1/3)� - (1/3)� + 0 =

3 * ((2/3)� - (1/3)�) = 31/81.