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Question Number 192098 by cortano12 last updated on 08/May/23
Answered by mehdee42 last updated on 08/May/23
$$\left({c}\right)\:\:\frac{\mathrm{6}}{\mathrm{7}}\:\:\checkmark \\ $$
Commented by Skabetix last updated on 08/May/23
$${How}\:{sir}\:\:? \\ $$$${Probability}\:{they}\:{are}\:{chosen}\::\:\frac{\mathrm{1}}{\mathrm{7}}×\frac{\mathrm{1}}{\mathrm{7}}=\frac{\mathrm{1}}{\mathrm{49}} \\ $$$$\Rightarrow{Probability}\:{they}\:{are}\:{NOT}\:{chosen}\::\:\mathrm{1}−\frac{\mathrm{1}}{\mathrm{49}} \\ $$$$=\frac{\mathrm{49}}{\mathrm{49}}−\frac{\mathrm{1}}{\mathrm{49}}=\frac{\mathrm{49}−\mathrm{1}}{\mathrm{49}}=\frac{\mathrm{48}}{\mathrm{49}}\Rightarrow\left({e}\right) \\ $$
Commented by mehdee42 last updated on 08/May/23
$${groups}\:{of}\:\:{mens}\:\::\:{s},{m}_{\mathrm{1}} ,{m}_{\mathrm{2}} ,{m}_{\mathrm{3}} ,..,{m}_{\mathrm{7}} \\ $$$${women}'{s}\:{group}\::\:{j},{w}_{\mathrm{1}} ,{w}_{\mathrm{2}} ,{w}_{\mathrm{3}} ,..,{w}_{\mathrm{7}} \\ $$$${n}\left({S}\right)=\mathrm{7}!\:\:\:\:\:\&\:\:\:{n}\left({A}\right)=\mathrm{7}!−\mathrm{6}!=\mathrm{6}×\mathrm{6}! \\ $$$$\Rightarrow{p}\left({A}\right)=\frac{\mathrm{6}}{\mathrm{7}} \\ $$