Another hypergeometric problem I can't figure out

mfdoom11

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In endurance horse racing, people over the age of 15 are called seniors and people 15 years or younger are called juniors. An endurance horse race consists of 25 seniors and 8 juniors. (Give answer as a fraction or a decimal out to at least 4 places. If your answer is very small use scientific notation for example 3.3421E-6.)
What is the probability that the top three finishers were:
(a) all seniors


(b) all juniors


(c) 2 seniors and 1 junior


(d) 1 senior and 2 juniors

I am really stuck with this concept, can you show the steps and how it was completed
 
In endurance horse racing, people over the age of 15 are called seniors and people 15 years or younger are called juniors. An endurance horse race consists of 25 seniors and 8 juniors. (Give answer as a fraction or a decimal out to at least 4 places. If your answer is very small use scientific notation for example 3.3421E-6.)
So there are a total of 25+ 8= 33 horses.
What is the probability that the top three finishers were:
(a) all seniors
The probability the first horse is a senior is 25/33. There are then 32 horses left, 24 of them seniors. The probability the second horse is also a senior is 24/32. There are then 31 horses left, 23 of them seniors. The probability the third horse is also a senior is 23/31. The probability all three finishers is (25/33)(24/32)(23/31)= 0.42, approximately. That's not "very small".

(b) all juniors
Try this yourself this the same way as above.


(c) 2 seniors and 1 junior
A little bit different. The probability the first horse is a senior is 25/33 and the probability the second horse is a senior is 24/32 as above. There are then 31 horses, 8 of them juniors. The probability the third horse is a junior is 8/31. The probability the first and second horses are seniors and the third is a junior ("senior, senior, junior") is (25/33)(24/32)(8/31). But "2 seniors and 1 junior" could also be "senior, junior, senior". The probability the first horse is a senior is 25/33 as above. Then there are 32 horses, 8 of them juniors, so the probability the second horse is a junior is 8/32. Then there are 31 horses, 24 of them seniors so the probability the third horse is a senior is 24/31. The probability of (senior, junior, senior) in that order is (25/33)(8/32)(24/31). You should see that this is really the same number- just with the numerators in different order. Similarly the probability of (junior, senior, senior) is also (8/33)(25/32)(24/31), again the same number. The probability of two seniors and one junior is 3(25/33)(24/32)(8/31).


(d) 1 senior and 2 juniors
Try this now, the same as in the previous problem.

I am really stuck with this concept, can you show the steps and how it was completed
 
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