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Question Number 93330 by Shakhzod last updated on 12/May/20

$$\mathrm{A}\mathrm{committee}\mathrm{of}\mathrm{five}\mathrm{is}\mathrm{to}\mathrm{be}\mathrm{chosen}\mathrm{from}$$$$\mathrm{a}\mathrm{group}\mathrm{of}9\mathrm{people}.\mathrm{The}\mathrm{probability}$$$$\mathrm{that}\mathrm{a}\mathrm{certain}\mathrm{married}\mathrm{couple}\mathrm{will}\mathrm{either}$$$$\mathrm{serve}\mathrm{together}\mathrm{or}\mathrm{not}\mathrm{at}\mathrm{all}\mathrm{is}$$

Answered by prakash jain last updated on 12/May/20

$$\mathrm{let}\mathrm{members}\mathrm{be}{p}_{1,}{p}_2,...,p_9$$$$\mathrm{Let}(p_1,p_2)\mathrm{be}\mathrm{the}\mathrm{married}\mathrm{couple}$$$$\mathrm{Comittee}\mathrm{with}\mathrm{married}\mathrm{couple}$$$$\mathrm{as}\mathrm{members}={}^7{\mathrm{C}}_3=35$$$$\mathrm{Comittee}\mathrm{with}\mathrm{both}\mathrm{are}\mathrm{not}$$$$\mathrm{memers}={}^7{\mathrm{C}}_5=21$$$$\mathrm{Total}\mathrm{options}\mathrm{for}\mathrm{comitees}={}^9{\mathrm{C}}_5=126$$$$\mathrm{required}\mathrm{probability}=\frac{35+21}{126}=\frac{4}{11}$$

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