Explanation:
We are choosing a number at random from 1 to 10. That means there are
10 equally likely outcomes:
{1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
For each question, we count how many of these numbers satisfy the condition, then divide by 10 (since probability = favorable outcomes / total outcomes).
Let’s go one by one:
1)
Multiple of 2 between 1 and 10:
2, 4, 6, 8, 10 → 5 numbers
Probability = 5/10 =
1/2
2)
Multiple of 3 between 1 and 10:
3, 6, 9 → 3 numbers
Probability = 3/10
3)
Odd number between 1 and 10:
1, 3, 5, 7, 9 → 5 numbers
Probability = 5/10 =
1/2
4)
Not selecting a multiple of 2 = selecting an odd number (same as #3)
So again: 5/10 =
1/2
5)
Not selecting a multiple of 3
Multiples of 3: 3, 6, 9 → 3 numbers
So not multiples of 3: 10 − 3 = 7 numbers
Probability = 7/10
6)
Not selecting a multiple of 2 or a multiple of 3
First, find numbers that *are* multiples of 2 or 3 (union), then subtract from 10.
Multiples of 2: {2, 4, 6, 8, 10}
Multiples of 3: {3, 6, 9}
Union = {2, 3, 4, 6, 8, 9, 10} → 7 numbers
(Notice 6 appears in both, so we don’t double-count.)
So numbers that are
not multiples of 2 or 3: 10 − 7 = 3 numbers
Which ones? Let's list all 1–10 and remove those above:
1 → not multiple of 2 or 3
✔
2 → multiple of 2
✘
3 → multiple of 3
✘
4 → multiple of 2
✘
5 → not
✔
6 → multiple of both
✘
7 → not
✔
8 → multiple of 2
✘
9 → multiple of 3
✘
10 → multiple of 2
✘
So only 1, 5, 7 → 3 numbers
Probability = 3/10
7)
Selecting a multiple of 2 or a multiple of 3
From above, union has 7 numbers → probability = 7/10
8)
Selecting a 3 or smaller
Numbers: 1, 2, 3 → 3 numbers
Probability = 3/10
9)
Selecting an 8 or greater
Numbers: 8, 9, 10 → 3 numbers
Probability = 3/10
10)
Selecting factors of 4 AND factors of 6
First, find factors of 4: numbers that divide 4 evenly → 1, 2, 4
Factors of 6: 1, 2, 3, 6
Intersection (common to both): 1 and 2
So favorable outcomes = 2
Probability = 2/10 =
1/5
Now let’s write all answers clearly:
1) 1/2
2) 3/10
3) 1/2
4) 1/2
5) 7/10
6) 3/10
7) 7/10
8) 3/10
9) 3/10
10) 1/5
Double-check #10:
- Does 1 divide both 4 and 6? Yes.
- Does 2 divide both? Yes (4 ÷ 2 = 2, 6 ÷ 2 = 3).
- Does 4 divide 6? No.
- Does 3 divide 4? No.
So only 1 and 2 — correct.
All others also verified.
Final Answer:
1) $\frac{1}{2}$
2) $\frac{3}{10}$
3) $\frac{1}{2}$
4) $\frac{1}{2}$
5) $\frac{7}{10}$
6) $\frac{3}{10}$
7) $\frac{7}{10}$
8) $\frac{3}{10}$
9) $\frac{3}{10}$
10) $\frac{1}{5}$
Parent Tip: Review the logic above to help your child master the concept of 6th grade probability worksheet.