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Category: Probability and Statistics

Selecting-a-random-star-is-a-1-15-chance-at-random-Lets-say-you-have-to-pick-a-second-random-star-that-is-next-to-it-Either-above-below-or-to-the-si

Question Number 4753 by FilupSmith last updated on 05/Mar/16 $$\ast\:\:\ast\:\:\ast\:\:\ast\:\:\ast \\ $$$$\ast\:\:\ast\:\:\ast\:\:\ast\:\:\ast \\ $$$$\ast\:\:\ast\:\:\ast\:\:\ast\:\:\ast \\ $$$$ \\ $$$$\mathrm{Selecting}\:\mathrm{a}\:\mathrm{random}\:\mathrm{star}\:\mathrm{is}\:\mathrm{a}\:\:\frac{\mathrm{1}}{\mathrm{15}}\:\mathrm{chance} \\ $$$$\mathrm{at}\:\mathrm{random}.\:\mathrm{Lets}\:\mathrm{say}\:\mathrm{you}\:\mathrm{have}\:\mathrm{to}\:\mathrm{pick} \\ $$$$\mathrm{a}\:\mathrm{second}\:\mathrm{random}\:\mathrm{star}\:\mathrm{that}\:\mathrm{is}\:\mathrm{next}\:\mathrm{to}\:\mathrm{it}. \\ $$$${Either}\:{above},\:{below},\:{or}\:{to}\:{the}\:{side}. \\…

Probability-Urn-I-contains-5-red-and-3-green-balls-Urn-II-contains-2-red-and-7-green-balls-One-balls-is-transferred-at-random-from-urn-I-to-urn-II-After-stirring-1-ball-is-chosen-from-urn-II-

Question Number 135352 by EDWIN88 last updated on 12/Mar/21 $$\underline{\mathrm{Probability}} \\ $$$$\mathrm{Urn}\:\mathrm{I}\:\mathrm{contains}\:\mathrm{5}\:\mathrm{red}\:\mathrm{and}\:\mathrm{3}\:\mathrm{green}\:\mathrm{balls}.\:\mathrm{Urn}\:\mathrm{II} \\ $$$$\mathrm{contains}\:\mathrm{2}\:\mathrm{red}\:\mathrm{and}\:\mathrm{7}\:\mathrm{green}\:\mathrm{balls}.\:\mathrm{One}\:\mathrm{balls} \\ $$$$\mathrm{is}\:\mathrm{transferred}\:\left(\mathrm{at}\:\mathrm{random}\right)\:\mathrm{from}\:\mathrm{urn}\:\mathrm{I}\:\mathrm{to}\:\mathrm{urn}\:\mathrm{II} \\ $$$$.\:\mathrm{After}\:\mathrm{stirring}\:,\:\mathrm{1}\:\mathrm{ball}\:\mathrm{is}\:\mathrm{chosen}\:\mathrm{from}\:\mathrm{urn}\:\mathrm{II}. \\ $$$$\mathrm{What}\:\mathrm{is}\:\mathrm{probability}\:\mathrm{that}\:\mathrm{the}\:\left(\mathrm{final}\right)\:\mathrm{ball}\:\mathrm{is} \\ $$$$\mathrm{green}?\: \\ $$ Answered…

Question-4166

Question Number 4166 by Filup last updated on 30/Dec/15 Commented by prakash jain last updated on 31/Dec/15 $$\mathrm{To}\:\mathrm{go}\:\mathrm{broke}:\:{k}\:\mathrm{wins},\:\mathrm{2}+{k}\:\mathrm{losses} \\ $$$$\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\left(\frac{\mathrm{3}}{\mathrm{4}}\right)^{{k}} \left(\frac{\mathrm{1}}{\mathrm{4}}\right)^{\mathrm{2}+{k}} =\:\frac{\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} }{\mathrm{1}−\frac{\mathrm{3}}{\mathrm{16}}}\:=\:\frac{\mathrm{1}}{\mathrm{13}}\:?…

A-bag-has-4-red-marbles-5-white-marbles-and-6-blue-marbles-Three-marbles-are-drawn-from-the-bag-without-replacement-What-is-the-probability-that-they-are-all-the-same-color-

Question Number 135181 by liberty last updated on 11/Mar/21 $$\mathrm{A}\:\mathrm{bag}\:\mathrm{has}\:\mathrm{4}\:\mathrm{red}\:\mathrm{marbles},\:\mathrm{5}\:\mathrm{white}\: \\ $$$$\mathrm{marbles}\:,\:\mathrm{and}\:\mathrm{6}\:\mathrm{blue}\:\mathrm{marbles}.\:\mathrm{Three} \\ $$$$\mathrm{marbles}\:\mathrm{are}\:\mathrm{drawn}\:\mathrm{from}\:\mathrm{the}\:\mathrm{bag},\:\left(\mathrm{without}\right. \\ $$$$\left.\mathrm{replacement}\right)\:\mathrm{What}\:\mathrm{is}\:\mathrm{the}\:\mathrm{probability} \\ $$$$\mathrm{that}\:\mathrm{they}\:\mathrm{are}\:\mathrm{all}\:\mathrm{the}\:\mathrm{same}\:\mathrm{color}\:?\: \\ $$ Answered by EDWIN88 last updated…

I-have-n-six-sided-dice-I-roll-them-all-What-is-the-proability-that-k-of-them-share-the-same-value-

Question Number 3894 by Filup last updated on 24/Dec/15 $$\mathrm{I}\:\mathrm{have}\:{n}\:\mathrm{six}\:\mathrm{sided}\:\mathrm{dice}. \\ $$$$\mathrm{I}\:\mathrm{roll}\:\mathrm{them}\:\mathrm{all}.\:\mathrm{What}\:\mathrm{is}\:\mathrm{the}\:\mathrm{proability} \\ $$$$\mathrm{that}\:{k}\:\mathrm{of}\:\mathrm{them}\:\mathrm{share}\:\mathrm{the}\:\mathrm{same}\:\mathrm{value}? \\ $$ Commented by Filup last updated on 24/Dec/15 $$\mathrm{Whoops}.\:\mathrm{That}\:\mathrm{was}\:\mathrm{my}\:\mathrm{mistake}! \\…

Box-I-has-3-red-and-5-white-balls-while-Box-II-contains-4-red-and-2-white-balls-A-ball-is-chosen-at-random-from-the-first-box-and-placed-in-the-second-box-without-observing-its-colour-Then-a-ball-

Question Number 3824 by Yozzii last updated on 21/Dec/15 $${Box}\:{I}\:{has}\:\mathrm{3}\:{red}\:{and}\:\mathrm{5}\:{white}\:{balls}, \\ $$$${while}\:{Box}\:{II}\:{contains}\:\mathrm{4}\:{red}\:{and}\:\mathrm{2}\: \\ $$$${white}\:{balls}.\:{A}\:{ball}\:{is}\:{chosen}\:{at}\:{random} \\ $$$${from}\:{the}\:{first}\:{box}\:{and}\:{placed}\:{in}\:{the} \\ $$$${second}\:{box}\:{without}\:{observing}\:{its}\:{colour}. \\ $$$${Then}\:{a}\:{ball}\:{is}\:{drawn}\:{from}\:{the}\:{second} \\ $$$${box}.\:{Find}\:{the}\:{probability}\:{that}\:{it}\:{is}\:{white}. \\ $$ Commented…