Question

production facility employs 10 workers on the day shift, 8 workers on the swing shift, and 6 workers on the graveyard shift. A quality control consultant is to select 5 of these workers for in-depth interviews. Suppose the selection is made in such a way that any particular group of 5 workers has the same chance of being selected as does any other group (drawing 5 slips without replacement from among 24). (a) How many selections result in all 5 workers coming from the day shift? selections What is the probability that all 5 selected workers will be from the day shift? (Round your answer to four decimal places.) (b) What is the probability that all 5 selected workers will be from the same shift? (Round your answer to four decimal places.) (c) What is the probability that at least two different shifts will be represented among the selected workers? (Round your answer to four decimal places.) (d) What is the probability that at least one of the shifts will be unrepresented in the sample of workers? (Round your answer to four decimal places.)

Answer 1

Step-by-step explanation:

A="All 5 selected workers will be from the day shift"

B="All 5 selected workers will be from the same shift"

C="At least two different shifts will be represented"

D="At least one of the shifts will be unrepresented in the sample of workers?"

a) #selections=$(^{10}_5)=\frac{10!}{5!(10-5)!}=252$

$P(A)=\frac{(^{10}_5)}{(^{24}_5)}}=\frac{252}{42504}=\frac{3}{506}=5.929\times10^{-3}$

b) $P(B)=\frac{(^{10}_5)+(^8_5)+(^6_5)}{(^{24}_5)}}=\frac{252+56+6}{42504}=7.387\times^{-3}$

c) P(C)=1-P(B)=0.9926

d) P(D)=1-P(D')=$1-\frac{(^{10}_3)(^8_1)(^6_1)+(^{10}_2)(^8_2)(^6_1)+(^{10}_2)(^8_1)(^6_2)+(^{10}_1)(^8_3)(^6_1)+(^{10}_1)(^8_2)(^6_2)+(^{10}_1)(^8_1)(^6_3)}{(^{24}_5)}=1-0.6559=0.3441$