Question Number 45645 by byaw last updated on 15/Oct/18 Answered by MJS last updated on 15/Oct/18 $$\left\{\underset{\mathrm{1}} {\mathrm{1}},\:\underset{\mathrm{2}} {\mathrm{2}},\:\underset{\mathrm{2}} {\mathrm{3}},\:\underset{\mathrm{3}} {\mathrm{4}},\:\underset{\mathrm{2}} {\mathrm{5}},\:\underset{\mathrm{4}} {\mathrm{6}},\:\underset{\mathrm{2}} {\mathrm{8}},\:\underset{\mathrm{1}} {\mathrm{9}},\:\underset{\mathrm{2}}…
Question Number 111163 by Aina Samuel Temidayo last updated on 02/Sep/20 $$\mathrm{A}\:\mathrm{bag}\:\mathrm{contains}\:\mathrm{11}\:\mathrm{white}\:\mathrm{balls}\:\mathrm{and}\:\mathrm{9} \\ $$$$\mathrm{black}\:\mathrm{balls},\:\mathrm{another}\:\mathrm{contains}\:\mathrm{12}\:\mathrm{white} \\ $$$$\mathrm{balls}\:\mathrm{and}\:\mathrm{13}\:\mathrm{black}\:\mathrm{balls}.\:\mathrm{If}\:\mathrm{two}\:\mathrm{balls} \\ $$$$\mathrm{are}\:\mathrm{drawn}\:\mathrm{without}\:\mathrm{replacement},\:\mathrm{from} \\ $$$$\mathrm{each}\:\mathrm{bag}\:\mathrm{find}\:\mathrm{the}\:\mathrm{probability}\:\mathrm{that} \\ $$$$\mathrm{exactly}\:\mathrm{one}\:\mathrm{of}\:\mathrm{the}\:\mathrm{4}\:\mathrm{balls}\:\mathrm{is}\:\mathrm{white}? \\ $$ Commented…
Question Number 110953 by bemath last updated on 01/Sep/20 Commented by Khanacademy last updated on 01/Sep/20 $$\boldsymbol{{Mumkin}}\:\:\boldsymbol{{bo}}'\boldsymbol{{lgan}}\:\:\boldsymbol{{imkoniyatlar}}\:\:\mathrm{9}\bullet\mathrm{8}/\mathrm{2}=\mathrm{36}. \\ $$$$\boldsymbol{{Ikkita}}\:\:\boldsymbol{{oqni}}\:\:\:\:\boldsymbol{{tanlash}}\:\:\:\:\boldsymbol{{mumkinligi}}\:\:\mathrm{5}\bullet\mathrm{4}/\mathrm{2}=\mathrm{10}.\:\: \\ $$$$\boldsymbol{{Demak}}:\:\:\:\mathrm{10}/\mathrm{36}=\mathrm{5}/\mathrm{18}\:\:\:\:\:\boldsymbol{{C}}. \\ $$$$\:\:\:\:\:\boldsymbol{{Telegram}}\:\:\:\boldsymbol{{manzil}}:\:\:\alpha\boldsymbol{{Quvonchbek}}\mathrm{3737} \\ $$…
Question Number 176449 by peter frank last updated on 19/Sep/22 $$\mathrm{A}\:\mathrm{fair}\:\mathrm{dice}\:\mathrm{was}\:\mathrm{thrown}\:\mathrm{twice}\:\mathrm{and} \\ $$$$\mathrm{it}\:\mathrm{landed}\:\mathrm{on}\:\mathrm{a}\:\mathrm{and}\:\mathrm{b}\:\mathrm{respectively}\:\mathrm{then} \\ $$$$\mathrm{the}\:\mathrm{probability}\:\mathrm{that}\:\mathrm{cubic}\:\mathrm{equation} \\ $$$$\mathrm{x}^{\mathrm{3}} −\left(\mathrm{3a}+\mathrm{1}\right)\mathrm{x}^{\mathrm{2}} +\left(\mathrm{3a}+\mathrm{2b}\right)\mathrm{x}−\mathrm{2b}=\mathrm{0} \\ $$$$\mathrm{has}\:\mathrm{three}\:\mathrm{distinct}\:\mathrm{root} \\ $$$$ \\ $$…
Question Number 110911 by Ar Brandon last updated on 31/Aug/20 $$\mathrm{Four}\:\mathrm{teachers}\:\mathrm{A},\:\mathrm{B},\:\mathrm{C},\:\mathrm{and}\:\mathrm{D}\:\mathrm{each}\:\mathrm{proposed}\:\mathrm{two}\:\mathrm{exercises}, \\ $$$$\mathrm{one}\:\mathrm{on}\:\mathrm{algebra}\:\mathrm{and}\:\mathrm{another}\:\mathrm{on}\:\mathrm{analyses},\:\mathrm{to}\:\mathrm{form}\:\mathrm{an}\:\mathrm{exam}. \\ $$$$\mathrm{The}\:\mathrm{students}\:\mathrm{have}\:\mathrm{to}\:\mathrm{choose}\:\mathrm{two}\:\mathrm{exercises}\:\mathrm{at}\:\mathrm{random}. \\ $$$$\mathrm{1}.\:\mathrm{Calculate}\:\mathrm{the}\:\mathrm{probability}\:\mathrm{P}\left(\mathrm{a}\right)\:\mathrm{of}\:\mathrm{a}\:\mathrm{student}\:\mathrm{to}\:\mathrm{choose} \\ $$$$\mathrm{two}\:\mathrm{exercises}\:\mathrm{on}\:\mathrm{algebra}. \\ $$$$\mathrm{a}\backslash\:\mathrm{P}\left(\mathrm{a}\right)=\frac{\mathrm{3}}{\mathrm{16}}\:,\:\mathrm{b}\backslash\mathrm{P}\left(\mathrm{a}\right)=\frac{\mathrm{3}}{\mathrm{14}}\:,\:\mathrm{c}\backslash\mathrm{P}\left(\mathrm{a}\right)=\frac{\mathrm{1}}{\mathrm{4}}\:,\:\mathrm{d}\backslash\mathrm{None} \\ $$$$\mathrm{2}.\:\mathrm{Calculate}\:\mathrm{the}\:\mathrm{probability}\:\mathrm{P}\left(\mathrm{b}\right)\:\mathrm{of}\:\mathrm{choosing}\:\mathrm{two}\:\mathrm{exercises} \\ $$$$\mathrm{proposed}\:\mathrm{by}\:\mathrm{the}\:\mathrm{same}\:\mathrm{teacher}.…
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Question Number 110826 by Aina Samuel Temidayo last updated on 30/Aug/20 $$\mathrm{The}\:\mathrm{probability}\:\mathrm{that}\:\mathrm{at}\:\mathrm{least}\:\mathrm{one}\:\mathrm{of} \\ $$$$\mathrm{the}\:\mathrm{events}\:\mathrm{A}\:\mathrm{and}\:\mathrm{B}\:\mathrm{occurs}\:\mathrm{is}\:\mathrm{0}.\mathrm{7}\:\mathrm{and} \\ $$$$\mathrm{they}\:\mathrm{occur}\:\mathrm{simultaneously}\:\mathrm{with} \\ $$$$\mathrm{probability}\:\mathrm{0}.\mathrm{2}.\:\mathrm{Then}\:\mathrm{P}\left(\overset{−} {\mathrm{A}}\right)+\mathrm{P}\left(\bar {\mathrm{B}}\right)\:= \\ $$ Commented by kaivan.ahmadi…
Question Number 176168 by nadovic last updated on 14/Sep/22 $$\mathrm{Two}\:\mathrm{fair}\:\mathrm{dice}\:\mathrm{are}\:\mathrm{rolled}\:\mathrm{once}.\:\mathrm{Let}\:{X} \\ $$$$\mathrm{be}\:\mathrm{the}\:\mathrm{random}\:\mathrm{variable}\:\mathrm{representing} \\ $$$$\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{the}\:\mathrm{numbers}\:\mathrm{that}\:\mathrm{show} \\ $$$$\mathrm{up}\:\mathrm{on}\:\mathrm{the}\:\mathrm{two}\:\mathrm{dice}.\:\mathrm{Find}\:{X}. \\ $$ Answered by aleks041103 last updated on 14/Sep/22…
Question Number 44389 by rahul 19 last updated on 28/Sep/18 $${Given}\:\mathrm{2}\:{events}\:{A}\:{and}\:{B}\:{such}\:{that} \\ $$$${P}\left({A}\right)=\frac{\mathrm{1}}{\mathrm{3}}\:,\:{P}\left({A}\cup{B}\right)=\frac{\mathrm{3}}{\mathrm{4}}\:{then} \\ $$$${find}\:{range}\:{of}\:{P}\left({B}\right)? \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
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