Question . A production facility employs 20 workers on the day shift, 15 workers on the swing shift, and 10 workers on the graveyard shift. A quality control consultant is to select 6 of these workers for in-depth interviews. Suppose the selection is made in such a way that any particular group of 6 workers has the same chance of being selected as does any other group (drawing 6 slips without replacement from among 45). (a) How many selections result in all 6 workers coming from the day shift? What is the probability that all 6 selected workers will be from the day shift? (b) What is the probability that all 6 selected workers will be from the same shift? (c) What is the probability that at least two different shifts will be represented among the selected workers? (d) What is the probability that at least one of the shifts will be unrepresented in the sample of workers?
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No. of ways 6 selention can be made from 45=45 C_{6}a)P(\text { all } 6 \text { coure trom day })=\frac{{ }^{20} C_{6}}{4 \mathrm{SC}_{6}}=4.759 \times 10^{-3}\text { Ausa } 4.759 \times 10^{-3}(b) P( all 6 come from 1 shift )=P(B)\frac{{ }^{20} C_{6}+{ }^{15} C_{6}+{ }^{10} C_{6}}{45 C_{6} .}=5.399 \times 10^{-3}c) Probability, that at least 2 shift are repeesented =1 - Probability that all come from 1 shift =1-0.0053991-P(B)=0.9946 \text { Ans } Cd) It A_{1} is the event that 1 st shift in upresentented A_{2} " " " 2^{\text {nd }} "A_{3} " " " " 3^{\text {rd }}\begin{array}{l}P\left(A_{1} \cup A_{2} \cup A_{3}\right)=P\left(A_{1}\right)+P\left(A_{2}\right)+P\left(A_{3}\right)-P\left(A_{1} \cap A_{2}\right) \\-P\left(A_{2} \cap A_{3}\right)-P\left(A_{1} \cap A_{3}\right)-P\left(A_{1} \cap A_{2} \cap A_{3}\right. \\=\frac{25 C_{6}}{45 C_{6}}+\frac{{ }^{30} C_{6}}{45 C_{6}}+\frac{35 C_{6}}{45 C_{6}}-P(B)-0+P\left(A ^ { S } \left\{A_{1} A^{\left.A_{2}\right)+P\left(A_{2} \cap A_{3}\right)}\right.\right. \\=0.3 \text { Ahs } d \\\end{array} ...