Random without Replacement Probability?

110 people are consulted by a doctor. 45 are females of whom 20 are sick and 25 are not.


The others are males of whom 15 are sick and the rest are not.


If 3 different people are selected at "random without replacement" what is the probability of:


a)- exactly two of the three being sick


b)- at least one is sick


c)- one of the three being a female is sick, one a male is sick, and one a female is NOT sick.

Random without Replacement Probability?
They are all combinatorics problems, so we are looking for the number of desirable outcomes, divided by the total number of possible outcomes. The number of possible outcomes is equal to (110 C 3) = 110!/(3!*107!) = 215,820





For a) we have the desirable outcomes choosing 2 sick people out of the 35 total and 1 health out of the 75 total so:


(35 C 2)*(75 C 1) = 44,625


Take that over 215,820 and we get .20676953





For b) we want the compliment of there being no sick people, which would be choosing 3 healthy people out of the 75


(75 C 3) = 67,525


Take that over 215,820 and we get .31287647


Subtract that from 1 to get our answer of .687123529





For c) we want to choose 1 sick female from the 20 possible, 1 sick male from the 15 possible and 1 helathy female from the 25 possible or (20 C 1)*(15 C 1)*(25 C 1) = 7,500


Take that over 215,820 to get our answer of .034751182
Reply:Sick = 20+15 = 35


Not sick = 25+(65-15) = 75





a)


P(sick,sick,not) or P(sick,not,sick) or P(not,sick,sick)


= (35/110)(34/109)(75/108) x 3


= 0.0689 (to 3 sig fig)





b)


P(at least 1 sick)


= 1 - P(none are sick)


= 1 - (75/110)(74/109)(73/108)


= 0.687 (to 3 sig fig)





c)


P(female sick, male sick, female not sick)


= (20/110)(15/109)(25/108) x 6


= 0.0348 (to 3 sig fig)


[The 6 is for the different combinations, as in a).]
Reply:a) How many ways can you select 3 different people (at random) out of 110 people?


Answer: 110C3, which equals 110! / (107! 3!) =


(110)(109)(108) / (3)(2)(1) = 215,820





How many of these will involve 2 sick people and 1 healthy person? In other words, how many ways are there to select 2 sick and one healthy?


Answer: (35C2)(75C1) = [(35)(34) / 2] [75] = 44,625





So the answer to a) is the second number divided by the first:


44,625 / 215,820 = .2067695, which is about 20.7%





b) The denominator will be the same (215,820), but the numerator is the number of ways of selecting 1, 2, or 3 sick people (and 2, 1, or 0 healthy people).





You can select 1 sick person in 35 ways (= 35C1), and you can select 2 healthy people in 75C2 ways (= (75)(74)/2 = 2,775. So there are 35 x 2,775 ways to choose exactly one sick person, which equals 97,125.





For 2 sick and one healthy, the numerator is (35C2)(75C1) = 44,625.





For 3 sick and none healthy, it is 35C3 = 6,545.





The numerator is the sum of these 3 numbers, and the denominator is again 215,820, so the probability is


(97,125 + 44,625 + 6,545)/ 215,820 = .687125, or 68.7%.





c)To choose one sick female, one sick male, and one healthy female, there are (20)(15)(25) = 7,500 ways, so the probability is 7,500 / 215,820 = .034751, or 3.5%.

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