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parveen110
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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).parveen110 wrote:Five distinguishable balls are placed into three distinguishable boxes. What is the probablity that atleast one 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].
