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kingwinner
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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!
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!
