# Probabilities involved in picking three numbers

• smckech
The "probability of picking two of the same numbers" is (1/10)(1/9)(1/8)= 1/720.In summary, the probability of picking three distinct numbers is 0.72, the probability of picking any three numbers in any order is 1, the probability of picking two of the same numbers is 1/10, and the probability of picking any two numbers in any order is 1/9.

#### smckech

There are three questions:

1. What is the probability of picking three distinct numbers (order does not matter)? (digits 0-9, ex: 123)

2. What is the probability of picking any three numbers in any order (order does not matter)? (digits 0-9, examples: 111,112,123)

3. What is the probability of picking two of the same numbers in any order (order does not matter)? (digits 0-9, examples 112,221,332)

My thoughts so far:

Question 1: I was thinking 10 c 3 and so it would be 120.

Question 2: The 'ask a scientist' website states that the answer is 120/1000. I disagree, I think it should be 220/1000 (a numerator greater than 120 at least, I may have got the 220 figure wrong).
My reasoning: Any three numbers could be three distinct, two of the same or all the same. Three distinct would be the 120, two of the same I think would be 90 and all the same would be 10. Giving a 220 total.

Question 3: This question is a part of question 2. I think it is 90 based on that fact that for each case of two numbers the same (10) there are 9 possibilities in each for the third number and so 9x10 = 90 possible ways of choosing two numbers the same.

A probability is a number between 0 and 1. All of your answers should be between 0 and 1.

Question 1: there are 10 c 3 ways of picking three distinct numbers, but what is the probability?

Question 2: this answer is just 1. It is always true, when picking three numbers, that 3 numbers are picked in some order.

My apologies, on rereading they were quite poorly phrased questions and I was mixing up questions.

If you don't mind I'll start afresh:

1. In a 3 ball lottery, where the ball is replaced after it is drawn, what is the probability of picking the correct number (the ordering does not matter, i.e. if you pick the same three numbers in any order then you win)?

I think it will depend on repeating numbers:

if you pick three distinct numbers P(x,y,z) = 3!/1000
if you pick two distinct numbers P(x,y,x) = 3/1000
if you pick three identical numbers P(x,x,x) = 1/1000

2. Under the same conditions, what is the probability the three numbers drawn are distinct?

3. Under the same conditions, what is the probability the three numbers have two repeats (xyx)?

4. Under the same conditions, what is the probability the three numbers drawn are the same?

5. As you correctly pointed out, the probability of any of the above three would be 1.

6. Lastly, if the balls are not replaced after being drawn then what is the probability of picking the correct number?

smckech said:
There are three questions:

1. What is the probability of picking three distinct numbers (order does not matter)? (digits 0-9, ex: 123)

2. What is the probability of picking any three numbers in any order (order does not matter)? (digits 0-9, examples: 111,112,123)

3. What is the probability of picking two of the same numbers in any order (order does not matter)? (digits 0-9, examples 112,221,332)

My thoughts so far:

Question 1: I was thinking 10 c 3 and so it would be 120.

Question 2: The 'ask a scientist' website states that the answer is 120/1000. I disagree, I think it should be 220/1000 (a numerator greater than 120 at least, I may have got the 220 figure wrong).
My reasoning: Any three numbers could be three distinct, two of the same or all the same. Three distinct would be the 120, two of the same I think would be 90 and all the same would be 10. Giving a 220 total.

Question 3: This question is a part of question 2. I think it is 90 based on that fact that for each case of two numbers the same (10) there are 9 possibilities in each for the third number and so 9x10 = 90 possible ways of choosing two numbers the same.