SOLUTION: I'm wrapping up a loooooong project for my Statistics class but there's just one problem that I can't figure out. I would be much obliged if anyone could help with it. It's 5 AM an
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Question 1131300: I'm wrapping up a loooooong project for my Statistics class but there's just one problem that I can't figure out. I would be much obliged if anyone could help with it. It's 5 AM and I'm exhausted. Brain no work no more. The project is due next Thursday but I have so much other crap to do in this class and my other classes that I actually went and did my Statistics homework early. I want to wrap up this project ASAP so I can move on to the other torture devices that await me.
Anyway....
Here goes:
The pulse rates of 101 sexy ladies were recorded. Here are all 101 pulse rates in ascending order: https://pastebin.com/JjWRG7RV
There are 31 different pulse rates among the 101 sexy ladies. The pulse rates and their absolute frequencies are listed here: https://pastebin.com/rhR4sYPs
Here is the question I don't know how to answer: "If five sexy ladies were selected out of the 101 sexy ladies, what is the probability that three of the five sexy ladies have pulse rates that equal 76 bpm?"
A pulse rate of 76 appears 8 times in the group of 101 pulse rates. My brain stops working here.
Please help.
Thank you. Answer by Boreal(15235) (Show Source):
You can put this solution on YOUR website! there are 10 ways (5C3) one can choose three from 5
the denominators of the following all multiply to the same number, and so do the numerators; they are just ordered differently, and there are 10 possible orders.
(8/101)(7/100)(6/99)(93/98)(92/97)*10=3259872/9505049400
=0.00034
