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

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

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