Please help (Probability problem)

[krisolaw]

Repairs: In how many orders can three broken computers and two broken printers be repaird if:

a. there are no restrictions (I get 5!=120)

b. the printers must be repaired first, and (book says answer is 12 so the only way I figure they got this one is 120/10 = 12)

c. the computers must be repaired first? (book says answer is 12 but how?)

d. If the order of repairs has no restrictions and the order of repairs is done at random, what is the probability that a printer will be repaired first?
(not really sure on this one either ~ book says answer is .4)

Please explain and show steps/work. Thanks in advance!

 

[pka]

(2!)(3!)=12: printers then computers.
Likewise (3!)(2!)=12: computers then printers.

 

[stapel]

You've been provided the solutions for (b) and (c).

For (d), you need to find the number of ways in which a printer could be chosen first. (Note: This is different from the situation in (b), where both printers were done first.)

So: In how many ways can you choose a printer to be the first thing chosen? And in how many ways can you order the remaining four units?

Now divide the number of ways of ordering the units so that "a printer" comes first by the total number of ways, found in (a), for ordering the units.

Hope that helps a bit.

Eliz.

 

[krisolaw]

Thanks Eliz! I got the last part figured out after pka showed me the other 2 steps. Thanks again!