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What is the probability of correctly choosing an unknown integer between 0 to 9 with three chances?

A. 963/1000 B. 973/1000 C. 983/1000 D. NONE

 

i think first  3 options are rubbish  ans should be 3/10 correct me if wrong

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First chance: Correct guessing in the first guessing.
So for the first time correct, we have The chance is 1/10.
Since we have guessed it correct in the first time, we don't need to count the second guessing and the third guessing.

Second chance: Correct guessing in the second guessing.
For the second time correct, we must have been guessed it wrong in the first chance, it's 9/10.
And for the correct guessing in the second guess, we only have 9 options left for X, since we will not choose the same number as we have guessed in the first guessing. So, it'll be 1/9.
So, the probability for this case is 9/10 * 1/9 = 1/10.

Third chance: We must have been guessed it wrong in the first and second guessing.
So, just like the previous case, the total probability for this case is 9/10 * 8/9 * 1/8 = 1/10

So, total probability is all cases = 1/10 + 1/10 + 1/10 = 3/10

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cant the answer be like this  ::

let an integer m exist in between 0<=m<=9 which is correct . so probability of choosing it would be 1/10 .

probability of choosing any other integer than m would be =9/10

so probability of choosing only wrong integers in 3 chances would be = (9/10)*(9/10)*(9/10)=729/1000

so probability of choosing the right integer would be = 1- 729/1000=271/1000 .

correct me if I am wrong .

Ans.D

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