Question 1202561: A panel of 6 members is to be formed from 5 doctors and 10nurses to include each profession. Find, correct to two decimal places, the probability of selecting the members to include, a) equal doctors and nurses. b) at most 2 nurses c) more doctors than nurses
Answer by greenestamps(13203)   (Show Source):
You can put this solution on YOUR website!
Since the problem asks for the probabilities of 3 different outcomes, we might as well get practice with the required calculations by finding the probabilities of all possible combinations of doctors and nurses. That gives us practice in working this kind of problem; and it gives us a chance to check our calculations, since the sum of the probabilities of all cases must be 1.
Since we are looking at all cases, we can simplify the calculations by counting the numbers of ways of getting each outcome, instead of calculating each probability.
The total number of ways of choosing 6 of the 15 doctors and nurses is, literally, "15 choose 6":
(1) 5 doctors, 1 nurse: we need to choose all 5 of the 5 doctors and 1 of the 10 nurses:
(2) 4 doctors, 2 nurses: 
(3) 3 doctors, 3 nurses: 
(4) 2 doctors, 4 nurses: 
(5) 1 doctor, 5 nurses: 
(6) 0 doctors, 6 nurses: 
CHECK these calculations... 10+225+1200+2100+1260+210 = 5005 --- correct
Now we can answer the questions.
Note the problem says the panel must contain both doctors and nurses, so the last case is not allowed. So the total number of ways of choosing a panel with both doctors and nurses is 5005-210 = 4795. That will be the denominator of each of our probability fractions.
a) equal numbers of doctors and nurses (3 of each)
The probability is 1200/4795.
b) at most 2 nurses
The probability is (10+225)/4795 = 325/4795.
c) more doctors than nurses
That means 3, 4, or 5 doctors, which means at most 2 nurses, so this is equivalent to case b)
The probability is again 325/4795.
I leave it to you to do the busy work of converting the probabilities to decimals.
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