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15 Jan 2021, 08:30
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C
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Difficulty:
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51% (02:39) correct
49%
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wrong
based on 98
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In a certain company, three typists service for four departments. If the departments each send a file to the typists at random, what is the probability that every typist will recieve at least one file?
A. 8/9
B. 64/81
C. 4/9
D. 16/81
E. 5/9
Are You Up For the Challenge: 700 Level Questions
A. 8/9
B. 64/81
C. 4/9
D. 16/81
E. 5/9
Are You Up For the Challenge: 700 Level Questions
15 Jan 2021, 22:21
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Probability = Favorable Outcomes / Total Outcomes
The first department can send a file in 3 ways
Similarly the 2nd , 3rd and 4th department each can send a file in 3 ways each.
Therefore Total outcomes = \(3^4\)
Favorable Outcomes: Assume the 1st typist gets 1 file, the second gets 1 of the remaining 3 files and the 3rd typist gets the last 2 files.
This is 4C1 * 3C1 * 2C2 = 4 * 3 * 1 = 12
This distribution can be done in 3 ways (1, 1, 2) (1, 2, 1) and (2, 1,1). So total favorable outcomes = 12 * 3 = 36
The required probability = 36/81 = 4/9
Option C
Arun Kumar
The first department can send a file in 3 ways
Similarly the 2nd , 3rd and 4th department each can send a file in 3 ways each.
Therefore Total outcomes = \(3^4\)
Favorable Outcomes: Assume the 1st typist gets 1 file, the second gets 1 of the remaining 3 files and the 3rd typist gets the last 2 files.
This is 4C1 * 3C1 * 2C2 = 4 * 3 * 1 = 12
This distribution can be done in 3 ways (1, 1, 2) (1, 2, 1) and (2, 1,1). So total favorable outcomes = 12 * 3 = 36
The required probability = 36/81 = 4/9
Option C
Arun Kumar
15 Jan 2021, 22:23
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There are \(4\) departments. So, the total number of files sent from each department is \(4\). There is \(3\) typist.
Total possible outcome \(=3*3*3*3=81\)
The possible way that each typist get at least 1 file to type is \((1,1,2),(1,2,1),(2,1,1).\)-->1
Let's consider anyone case i.e, for eg \((1,1,2)\). One typist is assigned \(2\) files whereas the other assigned \(1\) file each.
The number of ways it can be done for one case is \(3!=1*2*3=6\)
For all possible pairs as mentioned above (1), \(4\) files can be distributed among \(3\) typists is \(3!=6\)
The number of ways to assign the files so that every typist is assigned at least 1 file \(= 6*6 = 36\).
Required probability, \(P=\frac{36}{81}=\frac{4}{9}\)
Ans C
Total possible outcome \(=3*3*3*3=81\)
The possible way that each typist get at least 1 file to type is \((1,1,2),(1,2,1),(2,1,1).\)-->1
Let's consider anyone case i.e, for eg \((1,1,2)\). One typist is assigned \(2\) files whereas the other assigned \(1\) file each.
The number of ways it can be done for one case is \(3!=1*2*3=6\)
For all possible pairs as mentioned above (1), \(4\) files can be distributed among \(3\) typists is \(3!=6\)
The number of ways to assign the files so that every typist is assigned at least 1 file \(= 6*6 = 36\).
Required probability, \(P=\frac{36}{81}=\frac{4}{9}\)
Ans C
