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 3881 by Yozzii last updated on 23/Dec/15

Four integers are chosen at random from 0 to 9, inclusive. Find the probability that no more than 2 integers are the same.

$${Four}\:{integers}\:{are}\:{chosen}\:{at}\:{random} \\ $$$${from}\:\mathrm{0}\:{to}\:\mathrm{9},\:{inclusive}.\:{Find}\:{the} \\ $$$${probability}\:{that}\:{no}\:{more}\:{than} \\ $$$$\mathrm{2}\:{integers}\:{are}\:{the}\:{same}.\: \\ $$$$ \\ $$

Commented by RasheedSindhi last updated on 24/Dec/15

1^(st) integer can be chosen in 10 ways. All are successful,So probability is 1 2^(nd) integer can be chosen in 10 ways. Case−i: Both chosen integers are same 3^(rd) integer can be chosen in 9 ways. 4^(th) integer can be chosen in 9 ways. Case−i total ways:10^2 .9^2 =8100 −−−−−−−−− Case−ii Both integers are different 3^(rd) integer can be chosen in 10 ways. SubCase−(i):3^(rd) integer is same to 1^(st) ∣ 2^(nd) 4^(th) integer can be chosen in 9 ways. Total ways:10^3 .9=9000 ......Continue

$$\mathrm{1}^{{st}} \:{integer}\:{can}\:{be}\:{chosen}\:{in}\:\mathrm{10}\:{ways}. \\ $$$${All}\:{are}\:{successful},{So}\:{probability} \\ $$$${is}\:\mathrm{1} \\ $$$$\mathrm{2}^{{nd}} \:{integer}\:{can}\:{be}\:{chosen}\:{in}\:\mathrm{10}\:{ways}. \\ $$$$\:\:{Case}−{i}:\:{Both}\:{chosen}\:{integers}\:{are}\:{same} \\ $$$$\mathrm{3}^{{rd}} \:{integer}\:{can}\:{be}\:{chosen}\:{in}\:\mathrm{9}\:{ways}. \\ $$$$\mathrm{4}^{{th}} \:{integer}\:{can}\:{be}\:{chosen}\:{in}\:\mathrm{9}\:{ways}. \\ $$$$\:\:\:{Case}−{i}\:{total}\:{ways}:\mathrm{10}^{\mathrm{2}} .\mathrm{9}^{\mathrm{2}} =\mathrm{8100} \\ $$$$−−−−−−−−− \\ $$$$\:\:\:\:\:{Case}−{ii}\:{Both}\:{integers}\:{are}\:{different} \\ $$$$\mathrm{3}^{{rd}} \:{integer}\:{can}\:{be}\:{chosen}\:{in}\:\mathrm{10}\:{ways}. \\ $$$$\:\:\:\:{SubCase}−\left({i}\right):\mathrm{3}^{{rd}} {integer}\:{is}\:{same}\:{to}\:\mathrm{1}^{{st}} \:\mid\:\mathrm{2}^{{nd}} \\ $$$$\mathrm{4}^{{th}} \:{integer}\:{can}\:{be}\:{chosen}\:{in}\:\mathrm{9}\:{ways}. \\ $$$${Total}\:{ways}:\mathrm{10}^{\mathrm{3}} .\mathrm{9}=\mathrm{9000} \\ $$$$\:\:\:\:\:\:......{Continue} \\ $$

Commented by Rasheed Soomro last updated on 25/Dec/15

Easy−To−Understand Approach! Sorry That I couldn′t think of it!

$$\mathcal{E}{asy}−\mathcal{T}{o}−\mathcal{U}{nderstand}\:\mathcal{A}{pproach}! \\ $$$$\mathcal{S}{orry}\:\mathcal{T}{hat}\:{I}\:{couldn}'{t}\:{think}\:{of}\:{it}! \\ $$

Commented by prakash jain last updated on 24/Dec/15

10 numbers with all 4 digits same. 10×9×4 with 3 digits same=360 probabilitu=((10000−370)/(10000))=((963)/(1000)) From 0000−9999 all number are allowed except numbers with all 4 digits or 3 digits are same. number with say digit 5 repeated 3 times. 4×9=36 numbers 4−choosing place of 1 different digit 9−number of digits different than 5.

$$\mathrm{10}\:\mathrm{numbers}\:\mathrm{with}\:\mathrm{all}\:\mathrm{4}\:\mathrm{digits}\:\mathrm{same}. \\ $$$$\mathrm{10}×\mathrm{9}×\mathrm{4}\:\mathrm{with}\:\mathrm{3}\:\mathrm{digits}\:\mathrm{same}=\mathrm{360} \\ $$$$\mathrm{probabilitu}=\frac{\mathrm{10000}−\mathrm{370}}{\mathrm{10000}}=\frac{\mathrm{963}}{\mathrm{1000}} \\ $$$$\mathrm{From}\:\mathrm{0000}−\mathrm{9999}\:\mathrm{all}\:\mathrm{number}\:\mathrm{are}\:\mathrm{allowed} \\ $$$$\mathrm{except}\:\mathrm{numbers}\:\mathrm{with}\:\mathrm{all}\:\mathrm{4}\:\mathrm{digits}\:\mathrm{or}\:\mathrm{3}\:\mathrm{digits} \\ $$$$\mathrm{are}\:\mathrm{same}. \\ $$$$\mathrm{number}\:\mathrm{with}\:\mathrm{say}\:\mathrm{digit}\:\mathrm{5}\:\mathrm{repeated}\:\mathrm{3}\:\mathrm{times}. \\ $$$$\mathrm{4}×\mathrm{9}=\mathrm{36}\:\mathrm{numbers} \\ $$$$\mathrm{4}−\mathrm{choosing}\:\mathrm{place}\:\mathrm{of}\:\mathrm{1}\:\mathrm{different}\:\mathrm{digit} \\ $$$$\mathrm{9}−\mathrm{number}\:\mathrm{of}\:\mathrm{digits}\:\mathrm{different}\:\mathrm{than}\:\mathrm{5}. \\ $$

Commented by Yozzii last updated on 24/Dec/15

I understand now. Thanks!

$${I}\:{understand}\:{now}.\:{Thanks}! \\ $$

Answered by Rasheed Soomro last updated on 24/Dec/15

Let I_1 ,I_2 ,I_3 ,I_4 are integers ∈{0,1,2,...,9} I_1 (10ways)→I_2 (10ways) →I_3 { ((I_2 =I_1 9ways−I_4 9ways = 10^2 .9^2 =8100ways)),((I_2 ≠I_1 10ways−I_4 { ((I_3 ≠I_1 ∧ I_3 ≠I_2 10ways=10^4 =10000ways)),((I_3 =I_1 ∨ I_3 =I_2 9ways=10^3 .9=9000ways)) :})) :} Total ways:8100+10000+9000=27100

$$ \\ $$$${Let}\:{I}_{\mathrm{1}} ,{I}_{\mathrm{2}} ,{I}_{\mathrm{3}} ,{I}_{\mathrm{4}} \:{are}\:{integers}\:\in\left\{\mathrm{0},\mathrm{1},\mathrm{2},...,\mathrm{9}\right\} \\ $$$${I}_{\mathrm{1}} \left(\mathrm{10}{ways}\right)\rightarrow{I}_{\mathrm{2}} \left(\mathrm{10}{ways}\right) \\ $$$$\rightarrow{I}_{\mathrm{3}} \begin{cases}{{I}_{\mathrm{2}} ={I}_{\mathrm{1}} \:\mathrm{9}{ways}−{I}_{\mathrm{4}} \:\mathrm{9}{ways}\:=\:\mathrm{10}^{\mathrm{2}} .\mathrm{9}^{\mathrm{2}} =\mathrm{8100}{ways}}\\{{I}_{\mathrm{2}} \neq{I}_{\mathrm{1}} \:\mathrm{10}{ways}−{I}_{\mathrm{4}} \begin{cases}{{I}_{\mathrm{3}} \neq{I}_{\mathrm{1}} \:\wedge\:{I}_{\mathrm{3}} \neq{I}_{\mathrm{2}} \:\mathrm{10}{ways}=\mathrm{10}^{\mathrm{4}} =\mathrm{10000}{ways}}\\{{I}_{\mathrm{3}} ={I}_{\mathrm{1}} \:\vee\:{I}_{\mathrm{3}} ={I}_{\mathrm{2}} \:\mathrm{9}{ways}=\mathrm{10}^{\mathrm{3}} .\mathrm{9}=\mathrm{9000}{ways}}\end{cases}}\end{cases} \\ $$$$\mathcal{T}{otal}\:{ways}:\mathrm{8100}+\mathrm{10000}+\mathrm{9000}=\mathrm{27100} \\ $$$$\:\:\:\:\:\:\:\:\: \\ $$

Commented by prakash jain last updated on 24/Dec/15

I think total number of possibile outcomes are only 10000.

$$\mathrm{I}\:\mathrm{think}\:\mathrm{total}\:\mathrm{number}\:\mathrm{of}\:\mathrm{possibile}\:\mathrm{outcomes} \\ $$$$\mathrm{are}\:\mathrm{only}\:\mathrm{10000}. \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com