1. Suppose that there are 200 applicantsfor 5 positions at a certain company, an
ID: 2953592 • Letter: 1
Question
1. Suppose that there are 200 applicantsfor 5 positions at a certain company, and the company is able tonarrow the final field to 17 equally qualified candidates. Ten ofthe finalists are men and seven are women. Assume that the 5who are chosen for the positions are selected at random (not a goodway to hire workers!) from the final group of 17. Calculatethe probability that:
a. Four ofthe five hired are women.
b. Only twoof the males are hired.
c. At leastone of those hired is a male.
d. Exactlythree of the males are hired.
Explanation / Answer
Let Mi denote the event a male is selected Let Fi denote the even a female is selected Total males: 10 Total females: 7 5 positions to be filled a) P(4 of the 5 are women): (M1F2F3F4F5) + (F1M2F3F4F5) +(F1F2M3F4F5) + (F1F2F3M4F5) + (F1F2F3F4M5) (10/17)* (7/16) * 6/15 * 5/14 * 4/13 * 5 = 25/442 OR .056561 b)P(only two males): (M1M2F3F4F5) + (M1F2M3F4F5) + (M1F2F3M4F5) + (M1F2F3F4M5) + (F1M2M3F4F5) + (F1M2F3M4F5)+ (F1M2F3F4M5) + (F1F2M3M4F5) + (F1F2M3F4M5) + (F1F2F3M4M5)= (10/17)* (9/16) * (7/15) * (6/14) * (5/13) * 10 = 450/1768 OR.254525 c)P(at least one male): This problem has a shortcut of takingthe complement which is the probability of all females and findingthe diffrence between that and 1. The only reason you can dothis is because the sample spaceonly has one element that doesnt include at least one malewhich is all females. 1- P(F1F2F3F4F5) = 1 - (7/17 * 6/16 * 5/15 * 4/14 * 3/13) = 1 -.003394 = .996606 d)P(exactly three males): (M1M2M3F4F5) + (M1M2F3M4F5) +(M1M2F3F4M5) + (F1M2M3M4F5) +(F1M2M3F4M5) + (F1F2M3M4M5) (10/17)* 9/16 * 8/15 * 7/14 * 6/13 * 6 =54/221 OR .244344 HOPE THIS HELPSRelated Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.