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Am-not-a-friend-with-this-isssue-6-red-1-black-and-3-white-balls-We-draw-3-balls-in-a-raw-returning-the-drawn-ball-each-time-How-many-different-results-which-include-at-least-one-black-ball-W

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Question Number 178134 by Acem last updated on 13/Oct/22
Am not a friend with this isssue: 6 red, 1 black and 3 white balls We draw 3 balls in a raw, returning the drawn ball each time. How many different results which include at least one black ball. Way_1 : 1000−9^3 = 271 results That′s very ok Way_2 : 1×9×9_(very ok) ×3_(Omg, Why?) + 1×1×9×3_(the same confused above) + 1×1×1_(very ok and i like it.... but why not ×3) = 271 result Explanation_(about the story of ×3) ?
$${Am}\:{not}\:{a}\:{friend}\:{with}\:{this}\:{isssue}: \\ $$$$\mathrm{6}\:{red},\:\mathrm{1}\:{black}\:{and}\:\mathrm{3}\:{white}\:{balls} \\ $$$${We}\:{draw}\:\mathrm{3}\:{balls}\:{in}\:{a}\:{raw},\:{returning}\:{the} \\ $$$$\:{drawn}\:{ball}\:{each}\:{time}. \\ $$$${How}\:{many}\:{different}\:{results}\:{which}\:{include} \\ $$$$\:\boldsymbol{{at}}\:\boldsymbol{{least}}\:{one}\:{black}\:{ball}. \\ $$$$ \\ $$$$\:{Way}_{\mathrm{1}} :\:\mathrm{1000}−\mathrm{9}^{\mathrm{3}} =\:\mathrm{271}\:{results}\:{That}'{s}\:{very}\:{ok} \\ $$$$ \\ $$$$\:{Way}_{\mathrm{2}} :\:\mathrm{1}×\mathrm{9}×\mathrm{9}_{{very}\:{ok}} ×\mathrm{3}_{\boldsymbol{{Omg}},\:\boldsymbol{{Why}}?} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\:\mathrm{1}×\mathrm{1}×\mathrm{9}×\mathrm{3}_{{the}\:{same}\:{confused}\:{above}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\:\mathrm{1}×\mathrm{1}×\mathrm{1}_{{very}\:{ok}\:{and}\:{i}\:{like}\:{it}….\:{but}\:{why}\:{not}\:×\mathrm{3}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\mathrm{271}\:{result} \\ $$$$ \\ $$$$\:{Explanation}_{{about}\:{the}\:{story}\:{of}\:×\mathrm{3}} ? \\ $$$$ \\ $$
Commented by Acem last updated on 13/Oct/22
He asked me about final getting 1 black ball. I done! the question doesn′t care of arranging things. Anyway give me some time till Freon13 settle in my brain tank.
$$\:{He}\:{asked}\:{me}\:{about}\:{final}\:{getting}\:\mathrm{1}\:{black}\:{ball}. \\ $$$$\:{I}\:{done}!\:{the}\:{question}\:{doesn}'{t}\:{care} \\ $$$$\:{of}\:{arranging}\:{things}. \\ $$$$\:{Anyway}\:{give}\:{me}\:{some}\:{time}\:{till}\:{Freon}\mathrm{13} \\ $$$$\:{settle}\:{in}\:{my}\:{brain}\:{tank}.\: \\ $$$$ \\ $$
Commented by mr W last updated on 13/Oct/22
Explanation to Way 2 case 1: one black ball, two balls from other colors 1×9×9 since the black ball can be the first one, the second one or the third one, therefore totally 1×9×9×3 possibilities case 2: two black balls and one ball from other colors 1×1×9 since the black balls can be the first two, the last two or the first and the last one, therefore totally 1×1×9×3 possibilities case 3: three black balls 1×1×1 all together: 1×9×9×3+1×1×9×3+1×1×1=271
$${Explanation}\:{to}\:{Way}\:\mathrm{2} \\ $$$${case}\:\mathrm{1}:\:{one}\:{black}\:{ball},\:{two}\:{balls}\:{from} \\ $$$${other}\:{colors} \\ $$$$\mathrm{1}×\mathrm{9}×\mathrm{9} \\ $$$${since}\:{the}\:{black}\:{ball}\:{can}\:{be}\:{the}\:{first}\:{one}, \\ $$$${the}\:{second}\:{one}\:{or}\:{the}\:{third}\:{one}, \\ $$$${therefore}\:{totally}\:\mathrm{1}×\mathrm{9}×\mathrm{9}×\mathrm{3}\:{possibilities} \\ $$$${case}\:\mathrm{2}:\:{two}\:{black}\:{balls}\:{and}\:{one}\:{ball} \\ $$$${from}\:{other}\:{colors} \\ $$$$\mathrm{1}×\mathrm{1}×\mathrm{9} \\ $$$${since}\:{the}\:{black}\:{balls}\:{can}\:{be}\:{the}\:{first}\:{two}, \\ $$$${the}\:{last}\:{two}\:{or}\:{the}\:{first}\:{and}\:{the}\:{last} \\ $$$${one},\:\:{therefore}\:{totally}\:\mathrm{1}×\mathrm{1}×\mathrm{9}×\mathrm{3}\: \\ $$$${possibilities} \\ $$$${case}\:\mathrm{3}:\:{three}\:{black}\:{balls} \\ $$$$\mathrm{1}×\mathrm{1}×\mathrm{1} \\ $$$$ \\ $$$${all}\:{together}: \\ $$$$\mathrm{1}×\mathrm{9}×\mathrm{9}×\mathrm{3}+\mathrm{1}×\mathrm{1}×\mathrm{9}×\mathrm{3}+\mathrm{1}×\mathrm{1}×\mathrm{1}=\mathrm{271} \\ $$
Commented by mr W last updated on 13/Oct/22
if we had 2 black balls instead of one black ball in the bag, the result is 2×9×9×3+2×2×9×3+2×2×2=602 we can get this result also from 11^3 −9^3 =602
$${if}\:{we}\:{had}\:\mathrm{2}\:{black}\:{balls}\:{instead}\:{of}\:{one} \\ $$$${black}\:{ball}\:{in}\:{the}\:{bag},\:{the}\:{result}\:{is} \\ $$$$\mathrm{2}×\mathrm{9}×\mathrm{9}×\mathrm{3}+\mathrm{2}×\mathrm{2}×\mathrm{9}×\mathrm{3}+\mathrm{2}×\mathrm{2}×\mathrm{2}=\mathrm{602} \\ $$$${we}\:{can}\:{get}\:{this}\:{result}\:{also}\:{from} \\ $$$$\mathrm{11}^{\mathrm{3}} −\mathrm{9}^{\mathrm{3}} =\mathrm{602} \\ $$
Commented by Tawa11 last updated on 13/Oct/22
Great sir
$$\mathrm{Great}\:\mathrm{sir} \\ $$
Commented by Acem last updated on 13/Oct/22
First of all, thank you for your interest and detailed explanation. According to mathematical logic, we care about the order of the elements of the chosen set. For example choosing letters from a group to form a word (diff. words by changing the letters), choosing numbers to form numbers(diff. num.), choosing people to assign them specific tasks (diff. person′s job)...etc In all the above we concered in all their possibility permutations So how can you make me convince or accept mentally of the importance of the order of the selected only different color balls: • • • , • • • Here i see no difference, I can′t see that there′s different jobs, different phrases neither different numbers or even different forms. The result is i got one, two.. etc black balls According to way_1 : 10^3 −9^3 = 271 i know am wrong but not in all cases we can use the way_1 and that time will use the general principle counting then i′m sure i′ll make a mistake. Please help me to see that there are differences in the permutations at this issue. And make me friend with it.
$$ \\ $$$${First}\:{of}\:{all},\:{thank}\:{you}\:{for}\:{your}\:{interest}\:{and} \\ $$$$\:{detailed}\:{explanation}. \\ $$$$ \\ $$$${According}\:{to}\:{mathematical}\:{logic},\:{we}\:{care}\:{about} \\ $$$$\:{the}\:{order}\:{of}\:{the}\:{elements}\:{of}\:{the}\:{chosen}\:{set}. \\ $$$${For}\:{example}\:{choosing}\:{letters}\:{from}\:{a}\:{group}\:{to} \\ $$$${form}\:{a}\:{word}\:\left({diff}.\:{words}\:{by}\:{changing}\:{the}\:{letters}\right), \\ $$$${choosing}\:{numbers}\:{to}\:{form}\:{numbers}\left({diff}.\:{num}.\right), \\ $$$${choosing}\:{people}\:{to}\:{assign}\:{them} \\ $$$${specific}\:{tasks}\:\left({diff}.\:{person}'{s}\:{job}\right)…{etc} \\ $$$$ \\ $$$${In}\:{all}\:{the}\:{above}\:{we}\:{concered}\:{in}\:{all} \\ $$$$\:{their}\:\:{possibility}\:{permutations} \\ $$$$\:\boldsymbol{{So}}\:\boldsymbol{{how}}\:\boldsymbol{{can}}\:\boldsymbol{{you}}\:\boldsymbol{{make}}\:\boldsymbol{{me}}\:{convince}\:{or}\:{accept} \\ $$$$\:{mentally}\:{of}\:{the}\:{importance}\:{of}\:{the}\:{order}\:{of}\:{the} \\ $$$$\:{selected}\:{only}\:{different}\:{color}\:{balls}: \\ $$$$\:\bullet\:\bullet\:\bullet\:,\:\bullet\:\bullet\:\bullet\:{Here}\:{i}\:{see}\:{no}\:{difference},\:{I}\:{can}'{t} \\ $$$$\:{see}\:{that}\:{there}'{s}\:{different}\:{jobs},\:{different}\:{phrases} \\ $$$$\:{neither}\:{different}\:{numbers} \\ $$$$\:{or}\:{even}\:{different}\:{forms}. \\ $$$$ \\ $$$${The}\:{result}\:{is}\:{i}\:{got}\:{one},\:{two}..\:{etc}\:{black}\:{balls} \\ $$$$ \\ $$$$\:{According}\:{to}\:{way}_{\mathrm{1}} :\:\mathrm{10}^{\mathrm{3}} −\mathrm{9}^{\mathrm{3}} =\:\mathrm{271} \\ $$$$\:{i}\:{know}\:{am}\:{wrong}\:{but}\:{not}\:{in}\:{all}\:{cases}\:{we}\:{can}\:{use} \\ $$$$\:{the}\:{way}_{\mathrm{1}} \:{and}\:{that}\:{time}\:{will}\:{use}\:{the}\:{general} \\ $$$$\:{principle}\:{counting}\:{then}\:{i}'{m}\:{sure}\:{i}'{ll}\:{make} \\ $$$$\:{a}\:{mistake}. \\ $$$$ \\ $$$$\:{Please}\:{help}\:{me}\:{to}\:{see}\:{that}\:{there}\:{are}\:{differences} \\ $$$$\:{in}\:{the}\:{permutations}\:{at}\:{this}\:{issue}. \\ $$$$\:{And}\:{make}\:{me}\:{friend}\:{with}\:{it}. \\ $$$$ \\ $$$$ \\ $$$$ \\ $$
Commented by mr W last updated on 13/Oct/22
i can understand you doubt. even when we are not talking about probability in current question, but actually it is about probability. then we must treat the balls with the same color as different objects. example: in a bag there are 5 red balls and one black ball. now you randomly take two balls. how many possibilities for taking one red and one black ball? we know there are 5 possibilities, even they all look like the same: black+red. you don′t care which red ball is taken, because they are the same to you. but the balls themself care. there is difference if the ball A is taken or the ball B is taken.
$${i}\:{can}\:{understand}\:{you}\:{doubt}. \\ $$$${even}\:{when}\:{we}\:{are}\:{not}\:{talking}\:{about} \\ $$$${probability}\:{in}\:{current}\:{question},\:{but} \\ $$$${actually}\:{it}\:{is}\:{about}\:{probability}.\:{then} \\ $$$${we}\:{must}\:{treat}\:{the}\:{balls}\:{with}\:{the}\:{same} \\ $$$${color}\:{as}\:{different}\:{objects}.\:{example}: \\ $$$${in}\:{a}\:{bag}\:{there}\:{are}\:\mathrm{5}\:{red}\:{balls}\:{and}\:{one} \\ $$$${black}\:{ball}.\:{now}\:{you}\:{randomly}\:{take} \\ $$$${two}\:{balls}.\:{how}\:{many}\:{possibilities} \\ $$$${for}\:{taking}\:{one}\:{red}\:{and}\:{one}\:{black}\:{ball}? \\ $$$${we}\:{know}\:{there}\:{are}\:\mathrm{5}\:{possibilities}, \\ $$$${even}\:{they}\:{all}\:{look}\:{like}\:{the}\:{same}: \\ $$$${black}+{red}.\:{you}\:{don}'{t}\:{care}\:{which}\:{red}\: \\ $$$${ball}\:{is}\:{taken},\:{because}\:{they}\:{are}\:{the} \\ $$$${same}\:{to}\:{you}.\:{but}\:{the}\:{balls}\:{themself} \\ $$$${care}.\:{there}\:{is}\:{difference}\:{if}\:{the}\:{ball} \\ $$$${A}\:{is}\:{taken}\:{or}\:{the}\:{ball}\:{B}\:{is}\:{taken}. \\ $$
Commented by mr W last updated on 13/Oct/22
for the same reason in the world of probability • • • , • • • are two different things.
$${for}\:{the}\:{same}\:{reason}\:{in}\:{the}\:{world}\:{of} \\ $$$${probability} \\ $$$$\:\bullet\:\bullet\:\bullet\:,\:\bullet\:\bullet\:\bullet \\ $$$${are}\:{two}\:{different}\:{things}. \\ $$
Commented by Acem last updated on 13/Oct/22
I back to the qusetion, it asks about different results hmmm.... i need a bit time to see if there′s a difference between { ((how many result)),((how many different res.)) :}
$$\:{I}\:{back}\:{to}\:{the}\:{qusetion},\:{it}\:{asks}\:{about} \\ $$$$\:\:\boldsymbol{{different}}\:{results} \\ $$$$ \\ $$$${hmmm}….\:{i}\:{need}\:{a}\:{bit}\:{time}\:{to}\:{see}\:{if}\:{there}'{s} \\ $$$$\:{a}\:{difference}\:{between\begin{cases}{{how}\:{many}\:{result}}\\{{how}\:{many}\:{different}\:{res}.}\end{cases}} \\ $$$$ \\ $$

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