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I am working on this question:

An assembly operation for a computer circuit board consists of four operations that can be performed in any order.

a) In how many ways can the assembly operation be performed?

b) One of the operations involves soldering wire to a microchip. If all possible assembly orderings are equally likely, what is the probability that the soldering comes first or second?

In part a, I just use $4! = 24$

In part b, I can't get it, I'm a little confused about "comes first or second". Should I get the individual probabilities and just add them?

My answer in part b is $\frac{6}{64} + \frac{6}{64}$. Is that correct?

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Its rather unfortunate that you ask us 5 questions in a row in an hour or so. Please don't try to get your home work issues sorted out here. –  user21436 Jan 22 '12 at 14:35

1 Answer 1

Yes, you should get the individual probabilities and add them.

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I always like to see what answers get through the see-if-I'm-a-human-contributing-a-valuable-answer screen. This one (a generic agreement) did, although some short answers (but that actually have had content) of mine in the past haven't. How funny. –  mixedmath Jan 22 '12 at 14:28
    
my answer in letter b is 6/64 + 6/64. is that correct? –  IvanMatala Jan 22 '12 at 14:29
    
Let's see - how many total possibilities are there? There are $4! = 24$ total combinations, each equally likely. How many have soldering first? Choose soldering, then there are $3! = 6$. So the probability that soldering comes first is $6 / 24$. I don't think that's how you did it, right? –  mixedmath Jan 22 '12 at 14:32
    
and since each event are equal, the second event (which is to be the second) is 6/24 also? –  IvanMatala Jan 22 '12 at 14:36
    
@user: That's a good question. Are they really? –  mixedmath Jan 22 '12 at 14:41

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