Question and Answers Forum

All Questions      Topic List

Permutation and Combination Questions

Previous in All Question      Next in All Question      

Previous in Permutation and Combination      Next in Permutation and Combination      

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%

$$\mathrm{10}\:{different}\:{balls}\:{into}\:\mathrm{4}\:{identical}\: \\ $$$${groups}\:{with}\:\mathrm{2}\:{balls},\:\mathrm{2}\:{balls},\:\mathrm{3}\:{balls}, \\ $$$$\mathrm{3}\:{balls}\:{respectively}: \\ $$$$\frac{\mathrm{10}!}{\left(\mathrm{2}!\right)^{\mathrm{2}} \left(\mathrm{3}!\right)^{\mathrm{2}} \mathrm{2}!\mathrm{3}!}\:{ways} \\ $$$${since}\:{the}\:{boxes}\:{are}\:{different}, \\ $$$$\frac{\mathrm{10}!\mathrm{4}!}{\left(\mathrm{2}!\right)^{\mathrm{2}} \left(\mathrm{3}!\right)^{\mathrm{2}} \mathrm{2}!\mathrm{3}!}=\mathrm{50}\:\mathrm{400}\:{ways} \\ $$$${totally}\:\mathrm{4}^{\mathrm{10}} =\mathrm{1}\:\mathrm{048}\:\mathrm{576}\:{ways} \\ $$$$\Rightarrow{p}=\frac{\mathrm{50}\:\mathrm{400}}{\mathrm{1}\:\mathrm{048}\:\mathrm{576}}\approx\mathrm{4}.\mathrm{8\%} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com