# Simple Probability Problem

• s3a

## Homework Statement

PROBLEM(S):
Suppose that an operating room needs to schedule three knee, four hip, and five shoulder surgeries. Assume that all schedules are equally likely.

Determine the probability for each of the following:

(a) All hip surgeries are completed before another type of surgery.
(b) The schedule begins with a hip surgery.
(c) The first and last surgeries are hip surgeries.
(d) The first two surgeries are hip surgeries.

(a) 0.00202
(b) 1/3
(c) 1/11
(d) 1/11

## Homework Equations

I think set theory and baye's rule.

## The Attempt at a Solution

3/12 * 9/12 + 4/12 * 8/12 + 5/12 * 7/12

3/12 * 9/12 + 4/12 * 8/12 + 5/12 * 7/12
I'm sure that's supposed to mean something but what? What is your question?

Oh oops, for some reason, my question isn't here. :P

I meant to ask about part (a); I found out that the answer is 4/12 * 3/11 * 2/10 * 1/9. Having said that, my answer is equivalent to 1/(12C4) (read "1 over 12 choose 4"), but I don't know why. Could you please tell my why?

Oh oops, for some reason, my question isn't here. :P

I meant to ask about part (a); I found out that the answer is 4/12 * 3/11 * 2/10 * 1/9. Having said that, my answer is equivalent to 1/(12C4) (read "1 over 12 choose 4"), but I don't know why. Could you please tell my why?
I'm still not really understanding your question.

Just answer these two questions so that I can understand where you are in all this. What is the probability that the first operation will be a hip operation and then what is the probability that the second will be a hip operation?

What is 12C4 ?

12C4: http://www.wolframalpha.com/input/?i=12+choose+4

[1] What is the probability that the first operation will be a hip operation?
4/12

[2] What is the probability that the second will be a hip operation [assuming that the first operation was a hip operation]?
3/11

12C4: http://www.wolframalpha.com/input/?i=12+choose+4

[1] What is the probability that the first operation will be a hip operation?
4/12

[2] What is the probability that the second will be a hip operation [assuming that the first operation was a hip operation]?
3/11

Right. So you seem to understand the problem clearly so what is your question? I don't use wolframalpha and the page you linked to is not helping me understand what your question is.

Well, initially, I had no idea how to do (a), but, now, my issue is that I don't understand why my answer happens to equal to 1/("12 choose 4"). Apparently, it's not a coincidence, and I was hoping that you could provide me with the logic/reasoning as to why it is not a coincidence.

I agree it's not likely a coincidence but I'm not seeing the reason. I'm don't even see how "12 choose 4" gets to be 495 since when I do it I get 12*11*10*9 = 24 TIMES 495, not 495 but maybe they mean something different.

12 choose 4 = 12! / [(4!) * (12 - 4)!]

Also, I'm thinking that 12 choose 4 means "how many possible ways can there be 4 hip surgeries out of the 12 surgeries done in total?". Then, I suspect that 1/"12 choose 4" means "what is the probability that the aforementioned scenario happens once?".

I don't know if what I'm saying is correct, though.

12 choose 4 = 12! / [(4!) * (12 - 4)!]

Also, I'm thinking that 12 choose 4 means "how many possible ways can there be 4 hip surgeries out of the 12 surgeries done in total?". Then, I suspect that 1/"12 choose 4" means "what is the probability that the aforementioned scenario happens once?".

I don't know if what I'm saying is correct, though.
Happens once out of what? Once out of all the different possible orders the ops could be performed? . I get that as being 12*11*10*9 / 12! which is way lower than 1/495

What I was saying was that the event made up of the amount of ways 4 hip surgeries out of the 12 total surgeries can be done is the denominator, and that the probability is 1 of such scenarios out of all such scenarios, hence the taking of the inverse.

"12 choose 4" ... I get 12*11*10*9 = 24 TIMES 495
As s3a writes, 12 choose 4 is 12*11*10*9/(4*3*2*1)=495
"how many possible ways can there be 4 hip surgeries out of the 12 surgeries done in total?".
Yes, that works. There are 12 choose 4 ways the 4 hip surgeries can be allocated to the 12 slots. In only 1 of those do they take the first four slots.

As s3a writes, 12 choose 4 is 12*11*10*9/(4*3*2*1)=495
I'll be darned. Where is my logic wrong when I think 12 ways to choose the first, 11 to choose the second then 10 then 9? Why divide by anything?

I'll be darned. Where is my logic wrong when I think 12 ways to choose the first, 11 to choose the second then 10 then 9? Why divide by anything?
Because there are 24 different orders in which you could have chosen those four, but you don't care about their order.

Oh oops, for some reason, my question isn't here. :P

I meant to ask about part (a); I found out that the answer is 4/12 * 3/11 * 2/10 * 1/9. Having said that, my answer is equivalent to 1/(12C4) (read "1 over 12 choose 4"), but I don't know why. Could you please tell my why?

$$\frac{4}{12} \frac{3}{11} \frac{1}{10} \frac{1}{9}$$
$$\frac{1}{ \frac{12}{4} \frac{11}{3} \frac{10}{2} \frac{9}{1}}?$$
$$\frac{4}{12} \frac{3}{11} \frac{1}{10} \frac{1}{9}$$
$$\frac{1}{ \frac{12}{4} \frac{11}{3} \frac{10}{2} \frac{9}{1}}?$$