Question Number 106292 by bemath last updated on 04/Aug/20 $$\mathrm{a}\:\mathrm{box}\:\mathrm{contains}\:\mathrm{4}\:\mathrm{blue},\:\mathrm{3}\:\mathrm{green}\:\mathrm{and}\:\mathrm{2}\:\mathrm{red}\:\mathrm{identicall}\:\mathrm{balls}.\: \\ $$$$\mathrm{if}\:\mathrm{two}\:\mathrm{balls}\:\mathrm{are}\:\mathrm{selected}\:\mathrm{at}\:\mathrm{random}\:\mathrm{without}\: \\ $$$$\mathrm{replacement}\:,\:\mathrm{what}\:\mathrm{is}\:\mathrm{the}\:\mathrm{probability} \\ $$$$\mathrm{that}\:\mathrm{two}\:\mathrm{balls}\:\mathrm{be}\:\mathrm{of}\:\mathrm{the}\:\mathrm{same}\:\mathrm{colours}? \\ $$ Answered by bobhans last updated on 04/Aug/20…
Question Number 171761 by Mikenice last updated on 20/Jun/22 $${solve}: \\ $$$$\frac{{sin}\left(\mathrm{10}+{x}\right)}{{sin}\left(\mathrm{13}+{x}\right)}=\frac{{sin}\mathrm{27}{sin}\mathrm{39}}{{sin}\mathrm{24}{sin}\mathrm{57}} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 40550 by vajpaithegrate@gmail.com last updated on 24/Jul/18 $$\mathrm{if}\:\mathrm{three}\:\mathrm{numbers}\:\mathrm{are}\:\mathrm{drawn}\:\mathrm{at}\:\mathrm{random} \\ $$$$\mathrm{successively}\:\mathrm{without}\:\mathrm{replacement}\:\mathrm{from} \\ $$$$\mathrm{a}\:\mathrm{set}\:\mathrm{S}=\left\{\mathrm{1},\mathrm{2},……\mathrm{10}\right\}\mathrm{then}\:\mathrm{probability}\:\mathrm{that} \\ $$$$\mathrm{the}\:\mathrm{minimum}\:\mathrm{of}\:\mathrm{the}\:\mathrm{choosen}\:\mathrm{number}\:\mathrm{is} \\ $$$$\mathrm{3}\:\mathrm{or}\:\mathrm{their}\:\mathrm{maximum}\:\mathrm{is}\:\mathrm{7}\:\:\:\:\:\:\:\: \\ $$$$\:\mathrm{Answer}=\frac{\mathrm{11}}{\mathrm{40}} \\ $$ Answered by tanmay.chaudhury50@gmail.com…
Question Number 105994 by bemath last updated on 02/Aug/20 $$\mathbb{G}{iven}\:\begin{cases}{\mathrm{2log}\:_{\mathrm{5}} \left({x}+\mathrm{2}\right)−\mathrm{log}\:_{\mathrm{5}} \left({y}\right)+\frac{\mathrm{1}}{\mathrm{2}}=\mathrm{0}}\\{\mathrm{log}\:_{\mathrm{5}} \left({x}^{\mathrm{2}} +\mathrm{2}{x}−\mathrm{2}\right)=\mathrm{0}\:}\end{cases} \\ $$$${find}\:{the}\:{value}\:{of}\:{y}\: \\ $$ Commented by PRITHWISH SEN 2 last updated…
Question Number 105853 by bobhans last updated on 01/Aug/20 $${A}\:{box}\:{contains}\:\mathrm{5}\:{balls},\:\mathrm{2}\:{balls}\:{were} \\ $$$${drawn}\:{at}\:{random},\:{both}\:{of}\:{which} \\ $$$${turned}\:{out}\:{tobe}\:{white}.\:{what}\:{is}\:{the} \\ $$$${probability}\:{that}\:{all}\:{the}\:{balls}\:{in}\:{the}\:{box} \\ $$$${are}\:{white}\:? \\ $$ Commented by bobhans last updated…
Question Number 105791 by mr W last updated on 01/Aug/20 $${From}\:\mathrm{1}\:{to}\:\mathrm{12345},\:{how}\:{many}\:{numbers} \\ $$$${contain}\:{the}\:{digit}\:\mathrm{0}?\:{Find}\:{the}\:{number} \\ $$$${of}\:{zeros}\:{in}\:{all}\:{these}\:{numbers}. \\ $$$${Example}:\:\mathrm{10020}\:{has}\:{three}\:{zeros}. \\ $$ Commented by PRITHWISH SEN 2 last…
Question Number 40216 by scientist last updated on 17/Jul/18 Answered by math1967 last updated on 17/Jul/18 $$\overset{{n}} {\:}{c}_{{r}} +\overset{{n}} {\:}{c}_{{r}−\mathrm{1}} +\:^{{n}} {c}_{{r}−\mathrm{1}} +\overset{{n}} {\:}{c}_{{r}−\mathrm{2}} \\…
Question Number 171223 by mr W last updated on 07/Sep/22 $${In}\:{how}\:{many}\:{ways}\:{can}\:{you}\:{select}\:\mathrm{4} \\ $$$${from}\:\mathrm{40}\:{persons}\:{if}\:{every}\:{two}\:{persons} \\ $$$${may}\:{be}\:{selected}\:{together}\:{at}\:{most}\:{one}\: \\ $$$${time}? \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 105551 by mr W last updated on 29/Jul/20 $${In}\:{how}\:{many}\:{different}\:{ways}\:{can}\:\mathrm{10} \\ $$$${students}\:{be}\:{divided}\:{into}\:\mathrm{3}\:{groups}? \\ $$ Answered by adhigenz last updated on 29/Jul/20 $$\mathrm{It}\:\mathrm{is}\:\mathrm{the}\:\mathrm{same}\:\mathrm{thing}\:\mathrm{as}\:\mathrm{finding}\:\mathrm{how} \\ $$$$\mathrm{many}\:\mathrm{possible}\:\mathrm{integer}\:\mathrm{solution}\:\mathrm{for}…
Question Number 171012 by mr W last updated on 06/Jun/22 $$\mathrm{The}\:\mathrm{number}\:\mathrm{of}\:\mathrm{five}\:\mathrm{digits}\:\mathrm{can}\:\mathrm{be}\:\mathrm{made} \\ $$$$\mathrm{with}\:\mathrm{the}\:\mathrm{digits}\:\mathrm{1},\:\mathrm{2},\:\mathrm{3}\:\mathrm{each}\:\mathrm{of}\:\mathrm{which}\:\mathrm{can} \\ $$$$\mathrm{be}\:\mathrm{used}\:\mathrm{atmost}\:\mathrm{thrice}\:\mathrm{in}\:\mathrm{a}\:\mathrm{number}\:\mathrm{is} \\ $$ Commented by otchereabdullai@gmail.com last updated on 07/Jun/22 $$\mathrm{nice}\:\mathrm{question}\:+\:\mathrm{nice}\:\mathrm{solution}!…