Question 1150789: A store receives 5 red shirts and 10 green shirts. A random sample of 5 shirts selected. Determine the probability that
1. It contains 3 red shirts
2. It contain 1 red shirt.
3. It contain 3 green shirts
Found 2 solutions by ikleyn, greenestamps:
Answer by ikleyn(52792)   (Show Source):
You can put this solution on YOUR website! .
The total number of shirts is 5 + 10 = 15.
The answers are
1. P = (C[5]^3*C[10]^2])/C[15]^5.
2. P = (C[5]^1*C[10]^4])/C[15]^5.
3. P = (C[5]^2*C[10]^3])/C[15]^5.
The formulas are self-explanatory.
Answer by greenestamps(13200)  (Show Source):
You can put this solution on YOUR website!
Use the basic definition of probability:
number of favorable outcomes
P = ---------------------------------
total number of possible outcomes
For all three problems, the total number of possible outcomes is the number of ways of choosing 5 of the 15 shirts:  . So the denominator of the probability fraction for all three problems is 3003.
For the first problem, the favorable outcomes are the ones in which we choose 3 of the 5 red shirts AND 2 of the 10 green shirts: 
The probability is 450/3003. Simplify the fraction or convert to decimal if required.
The other two problems are exactly similar.
For the second problem the favorable outcomes have 1 of the 5 red shirts and 4 of the 10 green shirts: C(5,1)*C(10,4).
And for the third problem the favorable outcomes have 3 of the 10 green shirts and 2 of the 5 red shirts: C(10,3)*C(5,2).
You can finish the calculations for those last two problems....
If you want to get practice in making these kinds of calculations (that is, go beyond just answering the questions that were asked), find the probabilities for all the possible cases. The given problems have 3, 2, or 1 red shirts; perform similar calculations for the cases where there are 5, or 4, or 0 red shirts.
The sum of the probabilities for all possible cases should be 1; if not, you know you have made some calculation errors somewhere.
Note that, if you do this, it is by far the easiest if you leave all the fractions in unsimplified form; then you only need to verify that the numerators sum to 3003.
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