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If-10-different-balls-are-to-be-placed-in-4-boxes-at-random-then-the-probability-that-two-of-these-boxes-contain-exactly-2-and-3-balls-

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Question Number 181939 by greougoury555 last updated on 02/Dec/22
If 10 different balls are to be placed in 4 boxes at random , then the probability that two of these boxes contain exactly 2 and 3 balls
$${If}\:\mathrm{10}\:{different}\:{balls}\:{are}\:{to}\:{be}\:{placed} \\ $$$${in}\:\mathrm{4}\:{boxes}\:{at}\:{random}\:,\:{then}\:{the}\:{probability} \\ $$$${that}\:{two}\:{of}\:{these}\:{boxes}\:{contain} \\ $$$${exactly}\:\mathrm{2}\:{and}\:\mathrm{3}\:{balls}\: \\ $$
Answered by Acem last updated on 02/Dec/22
a• If you mean different boxes, here it is the solu. P(2B, 2,3 b.)= ((C_( 2) ^( 4) C_2 ^( 10) C_3 ^( 8) 2! 2^5 )/4^( 10) ) = ((3×2 × 45 × 2^4 ×7 ×2^5 )/2^(20) )= (( 945)/2^(10) ) b• if similar boxes, let me know, but the original is they are diff. because what it contain is diff.
$${a}\bullet\:{If}\:{you}\:{mean}\:{different}\:{boxes},\:{here}\:{it}\:{is}\:{the}\:{solu}. \\ $$$$ \\ $$$$\:{P}\left(\mathrm{2}{B},\:\mathrm{2},\mathrm{3}\:{b}.\right)=\:\frac{{C}_{\:\mathrm{2}} ^{\:\mathrm{4}} \:{C}_{\mathrm{2}} ^{\:\mathrm{10}} \:{C}_{\mathrm{3}} ^{\:\mathrm{8}} \:\mathrm{2}!\:\mathrm{2}^{\mathrm{5}} }{\mathrm{4}^{\:\mathrm{10}} } \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\frac{\mathrm{3}×\mathrm{2}\:×\:\mathrm{45}\:×\:\mathrm{2}^{\mathrm{4}} ×\mathrm{7}\:×\mathrm{2}^{\mathrm{5}} }{\mathrm{2}^{\mathrm{20}} }=\:\frac{\:\mathrm{945}}{\mathrm{2}^{\mathrm{10}} } \\ $$$$\:{b}\bullet\:{if}\:{similar}\:{boxes},\:{let}\:{me}\:{know},\:{but}\:{the}\:{original}\:{is}\:{they}\:{are}\: \\ $$$$\:\:\:\:\:\:{diff}.\:{because}\:{what}\:{it}\:{contain}\:{is}\:{diff}. \\ $$$$ \\ $$
Answered by mr W last updated on 03/Dec/22
to place 10 different balls into 4 different boxes there are totally 4^(10) ways. such that 2 boxes contain exactly 2 balls in each and 2 boxes contain exactly 3 balls in each, there are ((10!4!)/((2!)^3 (3!)^3 )) ways. p=((10!4!)/((2!)^3 (3!)^3 4^(10) ))=((1575)/(32 768))≈4.8%
$${to}\:{place}\:\mathrm{10}\:{different}\:{balls}\:{into}\:\mathrm{4}\: \\ $$$${different}\:{boxes}\:{there}\:{are}\:{totally} \\ $$$$\mathrm{4}^{\mathrm{10}} \:{ways}. \\ $$$${such}\:{that}\:\mathrm{2}\:{boxes}\:{contain}\:{exactly}\:\mathrm{2} \\ $$$${balls}\:{in}\:{each}\:{and}\:\mathrm{2}\:{boxes}\:{contain}\: \\ $$$${exactly}\:\mathrm{3}\:{balls}\:{in}\:{each},\:{there}\:{are}\: \\ $$$$\frac{\mathrm{10}!\mathrm{4}!}{\left(\mathrm{2}!\right)^{\mathrm{3}} \left(\mathrm{3}!\right)^{\mathrm{3}} }\:{ways}. \\ $$$${p}=\frac{\mathrm{10}!\mathrm{4}!}{\left(\mathrm{2}!\right)^{\mathrm{3}} \left(\mathrm{3}!\right)^{\mathrm{3}} \mathrm{4}^{\mathrm{10}} }=\frac{\mathrm{1575}}{\mathrm{32}\:\mathrm{768}}\approx\mathrm{4}.\mathrm{8\%} \\ $$
Commented by Acem last updated on 03/Dec/22
Hi Sir! According to the question, it asks about only box contains 2 balls, and the other contains 3 balls. right?
$${Hi}\:{Sir}! \\ $$$$\:{According}\:{to}\:{the}\:{question},\:{it}\:{asks}\:{about}\:{only}\:{box} \\ $$$$\:{contains}\:\mathrm{2}\:{balls},\:{and}\:{the}\:{other}\:{contains}\:\mathrm{3}\:{balls}. \\ $$$$\:{right}? \\ $$
Commented by mr W last updated on 03/Dec/22
the language of the question is not very clear. if it says “one box contains exactly 2 balls and an other box contains exactly 3 balls”, then it′s clear.
$${the}\:{language}\:{of}\:{the}\:{question}\:{is}\:{not}\: \\ $$$${very}\:{clear}.\:{if}\:{it}\:{says}\:“{one}\:{box}\:{contains} \\ $$$${exactly}\:\mathrm{2}\:{balls}\:{and}\:{an}\:{other}\:{box} \\ $$$${contains}\:{exactly}\:\mathrm{3}\:{balls}'',\:{then}\:{it}'{s} \\ $$$${clear}. \\ $$
Commented by Acem last updated on 03/Dec/22
No problem, but we got a new question! my solution is diffrent of yours please, let′s help each other • Which of the 2 boxes will contain 2 balls each? C_2 ^( 4) • Selecting the 2 first diff. balls C_( 2) ^( 10) now there are two methods to put them into_(B_1 , B_2 ) 2! _(I think that you won′t agree with me in this) • Selecting 2 other pairs of balls C_( 2) ^( 8) == • Selecting the rest C_3 ^( 6) C_3 ^( 3) 2! Num ways= C_2 ^( 4) C_( 2) ^( 10) C_( 2) ^( 8) 2! C_3 ^( 6) C_3 ^( 3) 2!= ((315)/2^( 6) ) P(A)= ((315)/2^( 14) )= 1.92 %
$${No}\:{problem},\:{but}\:{we}\:{got}\:{a}\:{new}\:{question}! \\ $$$$\:{my}\:{solution}\:{is}\:{diffrent}\:{of}\:{yours} \\ $$$$\:{please},\:{let}'{s}\:{help}\:{each}\:{other} \\ $$$$\:\bullet\:{Which}\:{of}\:{the}\:\mathrm{2}\:{boxes}\:{will}\:{contain}\:\mathrm{2}\:{balls}\:{each}? \\ $$$$\:\:\:\:\:{C}_{\mathrm{2}} ^{\:\mathrm{4}} \\ $$$$\:\bullet\:{Selecting}\:{the}\:\mathrm{2}\:{first}\:{diff}.\:{balls}\:{C}_{\:\mathrm{2}} ^{\:\mathrm{10}} \: \\ $$$$\:{now}\:{there}\:{are}\:{two}\:{methods}\:{to}\:{put}\:{them}\:{into}_{{B}_{\mathrm{1}} ,\:{B}_{\mathrm{2}} } \\ $$$$\:\:\:\:\:\:\mathrm{2}!\:_{{I}\:{think}\:{that}\:{you}\:{won}'{t}\:{agree}\:{with}\:{me}\:{in}\:{this}} \\ $$$$\:\bullet\:{Selecting}\:\mathrm{2}\:{other}\:{pairs}\:{of}\:{balls}\:{C}_{\:\mathrm{2}} ^{\:\mathrm{8}} \\ $$$$\:== \\ $$$$\:\bullet\:{Selecting}\:{the}\:{rest}\:{C}_{\mathrm{3}} ^{\:\mathrm{6}} {C}_{\mathrm{3}} ^{\:\mathrm{3}} \:\mathrm{2}! \\ $$$$\:{Num}\:{ways}=\:{C}_{\mathrm{2}} ^{\:\mathrm{4}} \:{C}_{\:\mathrm{2}} ^{\:\mathrm{10}} \:{C}_{\:\mathrm{2}} ^{\:\mathrm{8}} \:\mathrm{2}!\:\:{C}_{\mathrm{3}} ^{\:\mathrm{6}} {C}_{\mathrm{3}} ^{\:\mathrm{3}} \:\mathrm{2}!=\:\frac{\mathrm{315}}{\mathrm{2}^{\:\mathrm{6}} }\: \\ $$$$\:{P}\left({A}\right)=\:\frac{\mathrm{315}}{\mathrm{2}^{\:\mathrm{14}} }=\:\mathrm{1}.\mathrm{92}\:\%\: \\ $$
Commented by mr W last updated on 03/Dec/22
10 different balls into 4 identical groups with 2 balls, 2 balls, 3 balls, 3 balls respectively: ((10!)/((2!)^2 (3!)^2 2!3!)) ways since the boxes are different, ((10!4!)/((2!)^2 (3!)^2 2!3!))=50 400 ways totally 4^(10) =1 048 576 ways ⇒p=((50 400)/(1 048 576))≈4.8%
$$\mathrm{10}\:{different}\:{balls}\:{into}\:\mathrm{4}\:{identical}\: \\ $$$${groups}\:{with}\:\mathrm{2}\:{balls},\:\mathrm{2}\:{balls},\:\mathrm{3}\:{balls}, \\ $$$$\mathrm{3}\:{balls}\:{respectively}: \\ $$$$\frac{\mathrm{10}!}{\left(\mathrm{2}!\right)^{\mathrm{2}} \left(\mathrm{3}!\right)^{\mathrm{2}} \mathrm{2}!\mathrm{3}!}\:{ways} \\ $$$${since}\:{the}\:{boxes}\:{are}\:{different}, \\ $$$$\frac{\mathrm{10}!\mathrm{4}!}{\left(\mathrm{2}!\right)^{\mathrm{2}} \left(\mathrm{3}!\right)^{\mathrm{2}} \mathrm{2}!\mathrm{3}!}=\mathrm{50}\:\mathrm{400}\:{ways} \\ $$$${totally}\:\mathrm{4}^{\mathrm{10}} =\mathrm{1}\:\mathrm{048}\:\mathrm{576}\:{ways} \\ $$$$\Rightarrow{p}=\frac{\mathrm{50}\:\mathrm{400}}{\mathrm{1}\:\mathrm{048}\:\mathrm{576}}\approx\mathrm{4}.\mathrm{8\%} \\ $$

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