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What is the probability of choosing correctly an unknown integer between $0$ and $9$ with $3$ chances ?

1. $\frac{963}{1000}$
2. $\frac{973}{1000}$
3. $\frac{983}{1000}$
4. $\frac{953}{1000}$

I don't think any of the above options are correct
Answer should be  $\dfrac{271}{1000}$
prob of choosing correct integer = p(A)=1/10 , p(A)'= 9/10

prob of getting correct integer within 3 chances = 1/10( geeting correct in 1st chance )+ 9/10*1/10(within 2 chances)+9/10*9/10*1/10(within 3 chances )= 271/1000

none of the option

Lets find the probability of failure and subtract from 1 since we have 3 attempts.

In first attempt, probability of failure $= \frac{9}{10}.$

In second attempt, we won't select the number already tried. So, probability of failure $= \frac{8}{9}.$

Similarly, in third attempt, probability of failure $= \frac{7}{8}$.

So, probability of success $= 1 - \frac{9}{10}\frac{8}{9}\frac{7}{8} \\=1 - \frac{7}{10} = \frac{3}{10}.$

If after each chance, the corect number is changed, then the probability of failure for each try remains $\frac{9}{10}$ and we get probability of success $= 1 - {\frac{9}{10}}^3 = \frac{271}{1000}.$
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but question says with 3 chances  so it should have only one meaning that 3 chances with same number else it will imply 1 chance with 1 correct number          am i right ?
I think ans should be 3/10

p1+q1p2+ q1q2p3

Here p is probability of success and q is iprobability of failure and subscript represent attempt number...

(1/10) + (9/10) * (1/9) + (9/10) * (8/9) * (1/8)

So 3/10 should be ans..
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Ans None of these

Here we can get right digit in 1st attempt , 2nd attempt or 3rd attempt

Say,

Right number - R

Wrong number - W

So, attempt could be (R,_,_) ,(W,R,_) , (W,W,R)

So, probability is 1/10 +9/100 + 81/1000 =271/1000

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first chance 1/10 is ok

but why in 2nd chance 9/100  (it should be 9*1/10*9=1/10 as if a number is chosen then why someone again choose the same number ) and in 3rd 81/1000 (again it should be 9*8*1/10*9*8=1/10 by the same logic) so why not 3/10 or 300/1000 is the ans

p.s it is not a dice sort of problem where same number may come again
That depends on if the number is replaced after we give one choice.
Answers given in comment matches neither of the options.
Yes, such questions are called 'Marks to All' :)

If the unknown integer doesn't change in each chances

1.    when in first chance we got the integer

∴ Probability of getting the integer =$\dfrac{1}{10}$

1.   when in second chance we got the integer

∴ Probability of getting the integer =$\dfrac{9}{10} \times \dfrac{1}{10}$

1.   when in third chance we got the integer

∴ Probability of getting the integer =$\dfrac{9}{10} \times \dfrac{9}{10} \times \dfrac{1}{10}$

∴Probability of choosing correctly an unknown integer between 0 and 9 with 3 chances = $\dfrac{1}{10} + \left(\dfrac{9}{10} \times \dfrac{1}{10}\right) + \left(\dfrac{9}{10} \times \dfrac{9}{10} \times \dfrac{1}{10}\right)$

$=\dfrac{1}{10}+\dfrac{9}{100}+ \dfrac{81}{1000}$

$= \dfrac{271}{1000}$

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