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

$F o u r i n t e g e r s a r e c h o s e n a t r a n d o m$ $f r o m 0 t o 9, i n c l u s i v e . F i n d t h e$ $p r o b a b i l i t y t h a t n o m o r e t h a n$ $2 i n t e g e r s a r e t h e s a m e .$
Commented by RasheedSindhi last updated on 24/Dec/15

$1^{s t} i n t e g e r c a n b e c h o s e n i n 10 w a y s .$ $A l l a r e s u c c e s s f u l, S o p r o b a b i l i t y$ $i s 1$ $2^{n d} i n t e g e r c a n b e c h o s e n i n 10 w a y s .$ $C a s e - i : B o t h c h o s e n i n t e g e r s a r e s a m e$ $3^{r d} i n t e g e r c a n b e c h o s e n i n 9 w a y s .$ $4^{t h} i n t e g e r c a n b e c h o s e n i n 9 w a y s .$ $C a s e - i t o t a l w a y s : 10^{2} . 9^{2} = 8100$ $- - - - - - - - -$ $C a s e - i i B o t h i n t e g e r s a r e d i f f e r e n t$ $3^{r d} i n t e g e r c a n b e c h o s e n i n 10 w a y s .$ $S u b C a s e - (i) : 3^{r d} i n t e g e r i s s a m e t o 1^{s t} ∣ 2^{n d}$ $4^{t h} i n t e g e r c a n b e c h o s e n i n 9 w a y s .$ $T o t a l w a y s : 10^{3} . 9 = 9000$ $. . . . . . C o n t i n u e$
Commented by Rasheed Soomro last updated on 25/Dec/15

$E a s y - T o - U n d e r s t a n d A p p r o a c h!$ $S o r r y T h a t I {c o u l d n}^{'} t t h i n k o f i t!$
Commented by prakash jain last updated on 24/Dec/15

$10 numbers with all 4 digits same .$ $10 \times 9 \times 4 with 3 digits same = 360$ $probabilitu = \frac{10000 - 370}{10000} = \frac{963}{1000}$ $From 0000 - 9999 all number are allowed$ $except numbers with all 4 digits or 3 digits$ $are same .$ $number with say digit 5 repeated 3 times .$ $4 \times 9 = 36 numbers$ $4 - choosing place of 1 different digit$ $9 - number of digits different than 5 .$
Commented by Yozzii last updated on 24/Dec/15

$I u n d e r s t a n d n o w . T h a n k s!$
	Answered by Rasheed Soomro last updated on 24/Dec/15
	$Let I_1 ,I_2 ,I_3 ,I_4 are integers ∈{0,1,2,...,9} I_1 (10ways)→I_2 (10ways) →I_3 { ((I_2 =I_1 9ways−I_4 9ways = 10^2 .9^2 =8100ways)),((I_2 ≠I_1 10ways−I_4 { ((I_3 ≠I_1 ∧ I_3 ≠I_2 10ways=10^4 =10000ways)),((I_3 =I_1 ∨ I_3 =I_2 9ways=10^3 .9=9000ways)) :})) :} Total ways:8100+10000+9000=27100$
	$L e t I_{1}, I_{2}, I_{3}, I_{4} a r e i n t e g e r s \in {0, 1, 2, . . ., 9}$ $I_{1} (10 w a y s) \to I_{2} (10 w a y s)$ $\to I_{3} {\begin{cases} I_{2} = I_{1} 9 w a y s - I_{4} 9 w a y s = 10^{2} . 9^{2} = 8100 w a y s \\ I_{2} \neq I_{1} 10 w a y s - I_{4} {\begin{cases} I_{3} \neq I_{1} \land I_{3} \neq I_{2} 10 w a y s = 10^{4} = 10000 w a y s \\ I_{3} = I_{1} \lor I_{3} = I_{2} 9 w a y s = 10^{3} . 9 = 9000 w a y s \end{cases} \end{cases}$ $T o t a l w a y s : 8100 + 10000 + 9000 = 27100$
	Commented by prakash jain last updated on 24/Dec/15

	$I think total number of possibile outcomes$ $are only 10000 .$
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