1) In a lottery there are 10 tickets numbered 1,2,3,...,10. Two nubmers are drawn for prizes. You hold tickets 1 and 2. What is the probability that you win at least one prize? Method 1: assume order matters (different order=>different outcome) P(win at least one prize) = P(win exactly one prize) + P(win exactly two prizes) =P(1 is drawn but not 2) + P(2 is drawn but not 1) + P(both 1 and 2 are drawn) =[(1x8+8x1)+(1x8+8x1)+2] / (10x9) =17/45 Method 2: assume order does not matter P(win at least one prize) = P(win exactly one prize) + P(win exactly two prizes) =P(1 is drawn but not 2) + P(2 is drawn but not 1) + P(both 1 and 2 are drawn) =[(1x8)+(1x8)+1] / (10C2) =17/45 In this case, the probability is the same no matter we assume order is important or unimportant. I am just wondering...in general, is this always the case? If so, what is the actual reasoning behind it? 2a) A labor dispute has arisen concerning the distribution of 20 laborers to four different construction jobs. The first job (very undersirable) required 6 laborers; the second, third , and fourth utilized 4,5,and 5 laborers, respectively. The dispute arose over an alleged random distribution of the laborers to the jobs that placed all 4 members of a particular ethnic group on job 1. Find the probability of the observed event if it is assumed that the laborers are randomly assigned to jobs. 2b) Refer to part a. What is the probability that an ethnic group member is assigned to each type of job? 2c) Refer to part a. What is the probability that no ethnic group member is assigned to a type 4 job? 2a) my answer=[16!/(2!4!5!5!)] / [20!/(6!4!5!5!)] 2b) 2c) same denominator, but what would the numerator be? Can somebody please help me? Any help would be appreciated!
Regarding one. You either win at least one prize (P) or you win no prizes (Q). P + Q = 1 -- which is to say the odds of there being an outcome is 1 so the odds of all the possible outcomes must total to 1. Q = (8/10)(7/9) = 28/45. So P = 17 / 45 no matter how you go about it. I think the principle is P + Q = 1, so as long as you get all the cases of P or Q you know both P and Q.
Just relabel your ball "1" by "2" and vice versa. Do you think this should change any of the respective probabilities? Before I (or someone else) give you answers/help for the second, perhaps you could let us know what you've tried so far.
2b appears to start from the same principle. As for the question's statement, why does it imply the other 16 workers don't belong to ethnic groups? Please Register or Log in to view the hidden image!
1) So is it true in general that when calculating a "probability", (order unimportant) / (order unimportant) and (order important) / (order important) will always give the same answer? How to prove this rigorously? 2b) My guess is 4! [16!/(5!3!4!4!)] / [20!/(6!4!5!5!)], correct? 2c) Still unable to make a guess...
2c) I think the complement would be that "at least one ethnic group member is assigned to a type 4 job", wouldn't it be even more complicated then?
I don't think so. I haven't done these problems in a while so it might be an uninspired approach. Take the union of events where i ethnic members are assigned to a type 4 job, i =1,..,4. There is an expansion for the probability of that union (don't know the name for it).
2c) Is the answer 15/20 x 14/19 x 13/18 x 12/17 ??? How can I get the answer to part c using the approach (using factorials) as in parts a and b???
