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Question Number 81648 by Zainal Arifin last updated on 14/Feb/20

$$\mathrm{A}\mathrm{team}\mathrm{of}8\mathrm{couples},\left(\mathrm{husband}\mathrm{and}\mathrm{wife}\right)$$$$\mathrm{attend}\mathrm{a}\mathrm{lucky}\mathrm{draw}\mathrm{in}\mathrm{which}4\mathrm{persons}$$$$\mathrm{picked}\mathrm{up}\mathrm{for}\mathrm{a}\mathrm{prize}.\mathrm{Then}\mathrm{the}\mathrm{probability}$$$$\mathrm{that}\mathrm{there}\mathrm{is}\mathrm{at}\mathrm{least}\mathrm{one}\mathrm{couple}\mathrm{is}$$

Commented by mr W last updated on 14/Feb/20

$$probabilitythatatleastonecoupleis$$$$drawn:$$$$p=\frac{8\times C_2^{14}}{C_4^{16}}=\frac{8\times 91}{1820}=0.4=40\mathrm{\%}$$

Commented by mr W last updated on 14/Feb/20

$$probabilitythattwocouplesaredrawn:$$$$p=\frac{C_2^8}{C_4^{16}}=\frac{28}{1820}=1.54\mathrm{\%}$$

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