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Question Number 107659 by bobhans last updated on 12/Aug/20
$$\:\:\:\:\:\oint\mathbb{B}\mathrm{obhans}\oint \\ $$$$\mathrm{If}\:\mathrm{all}\:\mathrm{the}\:\mathrm{letters}\:\mathrm{of}\:\mathrm{the}\:\mathrm{word}\:\mathrm{MISSISSIPPI}\:\mathrm{are}\:\mathrm{written}\:\mathrm{down}\: \\ $$$$\mathrm{at}\:\mathrm{random}\:,\:\mathrm{the}\:\mathrm{probability}\:\mathrm{that}\:\mathrm{all}\:\mathrm{the}\:\mathrm{four}\:\mathrm{S}\:\mathrm{appear}\:\mathrm{consecutively}\:\mathrm{is}\:\_\_\_ \\ $$
Answered by bemath last updated on 12/Aug/20
$$\:\:\:\:\:\:\:\divideontimes\mathcal{B}{e}\mathcal{M}{ath}\divideontimes \\ $$$${n}\left({A}\right)=\boxdot\boxdot\boxdot\boxdot−−−−−−−=\:\mathrm{8}×\frac{\mathrm{7}!}{\mathrm{4}!.\mathrm{2}!} \\ $$$$\boxdot={letter}\:{of}\:{S} \\ $$$${n}\left({S}\right)=\:\frac{\mathrm{11}!}{\mathrm{4}!.\mathrm{4}!.\mathrm{2}!}\: \\ $$$${p}\left({A}\right)=\:\frac{\mathrm{8}.\mathrm{7}!}{\mathrm{4}!.\mathrm{2}!}\:×\:\frac{\mathrm{4}!.\mathrm{4}!.\mathrm{2}!}{\mathrm{11}!}=\:\frac{\mathrm{8}×\mathrm{24}×\mathrm{7}!}{\mathrm{11}.\mathrm{10}.\mathrm{9}.\mathrm{8}.\mathrm{7}!} \\ $$$$=\:\frac{\mathrm{24}}{\mathrm{11}.\mathrm{10}.\mathrm{9}}\:=\:\frac{\mathrm{4}}{\mathrm{11}.\mathrm{15}}=\:\frac{\mathrm{4}}{\mathrm{165}}\:. \\ $$
Commented by bobhans last updated on 12/Aug/20
$$\mathrm{yes}..\mathrm{correct} \\ $$