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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. $\frac{953}{1000}$
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I don't think any of the above options are correct
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Answer should be  $\dfrac{271}{1000}$
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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
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4 Answers

3 votes
3 votes
Best answer
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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1 comment

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 ?
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2 votes
2 votes
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..
0 votes
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
 

4 Comments

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
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That depends on if the number is replaced after we give one choice.
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Answers given in comment matches neither of the options.
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Yes, such questions are called 'Marks to All' :)
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0 votes
0 votes

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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