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.

Usernamevisible · Answer

Doubt -

The question says that three integers are "chosen" between 0 and 9, which to me suggests a combination, i.e., order does not matter. In that case, I would use 10C3 for the favorable cases. Since the denominator must represent all possible outcomes under the same selection process, shouldn't it be 10^3

thus giving 10c3/10^3

egmat · Answer

Hi Usernamevisible,

Your instinct about "chosen" isn't wrong, but here's the exact place it goes off track:  your numerator and denominator are counting two different kinds of things.

Look at what your own denominator represents. When you write  10^3 = 1000 , you're counting  ordered  triples - (2,6,7), (2,7,6), (6,2,7) are all sitting in there as separate outcomes. That's the only way the  1000  outcomes come out equally likely, which is what a probability ratio needs.

So the rule is simple:  whatever the denominator counts, the numerator must count the same way.

- Denominator counts ordered - numerator must count ordered.
-  10C3 = 120  counts  unordered  sets - it treats {2,6,7} as one thing and throws away its  6  arrangements.

Mixing them ( 120 / 1000 ) silently compares a bag of unordered sets against a pile of ordered triples. That's the inconsistency Bunuel was pointing at. To fix it, either multiply your numerator back up by 3! to make it ordered -  10C3 × 3! = 720  - or use the direct ordered count  10 × 9 × 8 = 720 . Either way:  720/1000 = 18/25 .

(Going fully unordered is a trap of its own, because unordered outcomes aren't equally likely - {2,2,5} happens less often than {2,6,7} - so you can't just divide counts. Ordered/ordered keeps everything clean.)

Lock it in with a tiny version

Pick  2  integers from just { 1 ,  2 ,  3 }, and ask: probability they're different?

- Ordered total:  3  ×  3  =  9  outcomes.
- Ordered favorable:  3  ×  2  =  6 . Probability =  6/9 = 2/3 .

Now try your mixed way:  3C2 = 3  on top,  9  on the bottom -  3/9 = 1/3 . You can list all  9  pairs by hand and count the "different" ones - you'll find  6 , not  3 . The unordered numerator dropped every reversed pair. Same slip, just small enough to see.

Answer: A

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