# Thread: Combinatorics and Probability - "3 Secys, 4 Depts"

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## Combinatorics and Probability - "3 Secys, 4 Depts"

There are three secretaries who work for four departments. If each of the four departments has one report to be typed out, and the reports are randomly assigned to a secretary, what is the probability that all three secretaries are assigned at least one report?

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Originally Posted by budablasta
There are three secretaries who work for four departments. If each of the four departments has one report to be typed out, and the reports are randomly assigned to a secretary, what is the probability that all three secretaries are assigned at least one report?
Let's see how the reports can be distributed...

1. ALL reports assigned to one secretary (0,0,4)

3!/2! = 3 ways

2. Three reports assigned to one secretary (0,1,3)

3! = 6 ways

3. Two reports assigned to one secretary (1,1,2)

3!/2! = 3 ways

Therefore, probability of at least one report per secretary is 3/9 = 1/3

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It's 4/9.

There are 3^4 total possibilities.

There are exactly three possibilities with a 4-0-0 distribution: 4C4*3.

There are 18 possibilities with a 2-2-0 distribution: 4C2*3

There are 24 possibilities with a 3-1-0 distribution: 4C3*3!

Finally, the only one that matters for the problem, there are 36 possibilities with a 2-1-1 distribution: 4C2*3!

36/81=4/9.

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Hi can anyone tell me how to solve this

A certain deck of cards contains 2 blue cards, 2 red cards, 2 yellow cards, and 2 green cards. If two cards are randomly drawn from the deck, what is the probability that they will both are not blue?

A. 15/28
B. 1/4
C. 9/16
D. 1/32
E. 1/16

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Here is how find the probability that neither card is blue:

If we're choosing two cards, there is a 6/8 chance that the first is not blue (because 2 of the 8 cards are blue) and then a 5/7 chance that the second card is not blue (because there are now only 7 cards to choose from and 5 are not blue).

This gives us a probability of (6/8)*(5/7) = 15/28