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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}.$

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}.$

0 votes

**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

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

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

1

0 votes

If the unknown integer doesn't change in each chances

- when in first chance we got the integer

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

- when in second chance we got the integer

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

- 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) $

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

$= \dfrac{271}{1000}$