Three integers are at random chosen between 0 and 9 inclusive. What is

Question

Three integers are at random chosen between 0 and 9 inclusive. What is the probability that each number is different ?

A. 18/25
B. 4/5
C. 81/100
D. 9/10
E. 1

Kurtosis · Accepted Answer

Number of ways of choosing the 3 integers without restriction = 10 * 10 * 10 
Number of ways of choosing the 3 integers so that the 3 integers are different = 10 * 9 * 8

Probability = 10 * 9 * 8/10 * 10 * 10 = 72/100 = 18/25

Answer: A

Funsho84 · Answer

First selection can be any 10 numbers, inclusive: 10/10 = 1
2nd selection will be reduced by 1 (because it has to be different from the 1st selection) : 9/10
3rd selection will be reduced by 2 (because it has to be different from selection 1 and selection 2): 8/10

1* (9/10) * (8/10) = 72/100
reduces down to 18/25

Answer A

arhumsid · Answer

(Prob. of choosing any one from 10) * (Prob. of not choosing the first one) * (Prob. of not choosing the first two)
 1 * (1-\frac{1}{10}) * (1-\frac{2}{10}) 
=> 1 * \frac{9}{10} * \frac{8}{10} 
=> \frac{72}{100} 
=> \frac{18}{25}

Option  A

akadiyan · Answer

Step 1: Probability of selecting 3 integers without constraint - 10*10*10
Total number of possible outcomes = 1000

Step 2: 
The probability of selecting first integer = 10
The probability of selecting second integer = 9 (Excluding the first digit)
The probability of selecting third integer = 8 (Excluding first and second digits)

Number of ways of selecting 3 different integers = 10*9*8 = 720

Total probability = 720/1000 = 72/100 = 18/25

Ans:   A

ScottTargetTestPrep · Answer

Let’s choose the first integer. . The probability that the second integer is different (from the first) is 9/10. The probability that the third integer is different (from the first and second) is 8/10.

Therefore, the probability that each number is different is 9/10 x 8/10 = 72/100 = 18/25.

Answer: A

ScottTargetTestPrep · Answer

The probability that the first number is different is 10/10 = 1 (since no other numbers have been chosen yet). The probability that the second number is different from the first is 9/10, and the probability the third number is different from the first two is 8/10 = 4/5. Therefore, the probability that all three numbers are different is 1 x 9/10 x 4/5 = 36/50 = 18/25.

Answer: A

Adarsh_24 · Answer

Bunuel

How would you solve this using combinations.

Here for the numbers chosen order doesnt matter right?
So when I pick 3 numbers, should it be 10C3 ?
Total possible will 10c1*10c1*10c1/3! Since there are repettition here too.
and both 3! cancel out in final probability?

Bunuel · Answer

The total number of ways to choose 3 integers from 10, allowing repetition, is 10 * 10 * 10 = 10^3 (as you've written, 10C1 * 10C1 * 10C1).

The number of ways to choose 3 different integers from 10, when order matters, is 10P3 = 720 (or 10C3 * 3!).

Thus, the probability of selecting 3 different integers is 720/1000 = 18/25.

Answer: A.

Adarsh_24 · Answer

Sorry, this is my exact doubt.
How come you consider order matters here?

I get the feeling that numbers are picked in a bunch when I read the question.

So 2,6,7 would be same as 2,7,6

Bunuel · Answer

You can think about it this way: The denominator, 10^3, consists of all possible three-digit sequences, meaning it represents  ordered  triplets. To maintain consistency, the numerator must also account for groups of three where order matters.

The key is to ensure consistency between the numerator and the denominator. If one considers ordered groups, the other must as well, and if one uses unordered groups, the other should too.

Adarsh_24 · Answer

Thank you for your insights as always, that does help.

Bharath.R · Answer

Hey!

Thank you for your answer, But I fail to understand why do we consider 2nd second and 3rd selection out of 10?(9/10,8/10) Shouldn't it be out of 9 and 8 respectively as we cannot choose the same integer again.

Bunuel · Answer

In this question, we are selecting integers from 0 to 9, inclusive. We are not selecting from physical slips of paper, so repeated numbers are allowed, meaning each selection is made from the same set of 10 possible integers. Please review the solutions above.

Isha2701 · Answer

Bunuel : I did it like Probability of selecting diff I= 1- prob(all same I)
And prob of all same I is: 1/10 x 1 x1 = 1/10 I wrote 1 twice because once we select one number out of 10, in order for it to be the same 2 more times, the prob is 1.
Hence, of selecting diff is= 1-1/10= 9/10. Where am I getting it wrong?

Bunuel · Answer

You are subtracting the wrong complement. “All different” is not the opposite of “all same.” There is a third case: exactly two numbers are the same, such as 2, 2, 5.

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