Question

total=16

1. what is the probability of spinning a sum of 3? (write in simple form)
2. what is the probability of spinning a sum of 5? (write in simple form)
3. what is the probability of spinning a sum of 3 or 7? (write in simple form)
4. what is the probability of spinning a sum less than 5? (write in simple form)

Explanation:

Step1: Identify total outcomes

Total possible outcomes = $16$

Step2: Find favorable for sum=3

Favorable pairs: (1,2), (2,1) → Count = $2$
Probability: $\frac{2}{16} = \frac{1}{8}$

Step3: Find favorable for sum=5

Favorable pairs: (1,4), (2,3), (3,2), (4,1) → Count = $4$
Probability: $\frac{4}{16} = \frac{1}{4}$

Step4: Find favorable for sum=3 or 7

Favorable for sum=7: (3,4), (4,3), (2,5) is invalid, assuming 1-4 spinners: (3,4), (4,3) → Count = $2$. Total favorable = $2+2=4$
Probability: $\frac{4}{16} = \frac{1}{4}$? Correction: Wait, standard 2x4 spinners: sum 3 has 2, sum7 has (3,4),(4,3) → 2. Total 4. $\frac{4}{16}=\frac{1}{4}$? No, wait, maybe spinners are 1-4 each: all pairs:
(1,1)=2, (1,2)=3, (1,3)=4, (1,4)=5
(2,1)=3, (2,2)=4, (2,3)=5, (2,4)=6
(3,1)=4, (3,2)=5, (3,3)=6, (3,4)=7
(4,1)=5, (4,2)=6, (4,3)=7, (4,4)=8
Sum 3: 2 outcomes, sum7: 2 outcomes. Total 4. $\frac{4}{16}=\frac{1}{4}$? Wait no, user's total is 16, so yes.

Step5: Find favorable for sum<5

Sum <5 means sum=2,3,4.
Sum2: (1,1) →1; sum3:2; sum4: (1,3),(2,2),(3,1)→3. Total favorable=1+2+3=6
Probability: $\frac{6}{16} = \frac{3}{8}$? Wait no, wait sum<5 is sum=2,3,4. Count:
Sum2:1, sum3:2, sum4:3. Total 6. $\frac{6}{16}=\frac{3}{8}$.

Wait correction:

1. Sum 3: 2 outcomes → $\frac{2}{16}=\frac{1}{8}$
2. Sum5: 4 outcomes → $\frac{4}{16}=\frac{1}{4}$
3. Sum3 or7: 2+2=4 → $\frac{4}{16}=\frac{1}{4}$
4. Sum<5: sum2(1)+sum3(2)+sum4(3)=6 → $\frac{6}{16}=\frac{3}{8}$

Wait I made a mistake earlier. Let's recheck:
All 16 pairs:

1. (1,1)=2
2. (1,2)=3
3. (1,3)=4
4. (1,4)=5
5. (2,1)=3
6. (2,2)=4
7. (2,3)=5
8. (2,4)=6
9. (3,1)=4
10. (3,2)=5
11. (3,3)=6
12. (3,4)=7
13. (4,1)=5
14. (4,2)=6
15. (4,3)=7
16. (4,4)=8

1. Sum3: outcomes 2 and5 → 2 → $\frac{2}{16}=\frac{1}{8}$
2. Sum5: outcomes4,7,10,13 →4 → $\frac{4}{16}=\frac{1}{4}$
3. Sum3 or7: sum3(2) + sum7(12,15)=2 → total 4 → $\frac{4}{16}=\frac{1}{4}$
4. Sum<5: sum2(1), sum3(2,5), sum4(3,6,9) →1+2+3=6 → $\frac{6}{16}=\frac{3}{8}$

Yes, that's correct.

Step1: Total possible outcomes

Total = $16$

Step2: Probability for sum=3

Favorable count = $2$
$\text{Probability} = \frac{2}{16} = \frac{1}{8}$

Step3: Probability for sum=5

Favorable count = $4$
$\text{Probability} = \frac{4}{16} = \frac{1}{4}$

Step4: Probability for sum=3 or7

Favorable count = $2+2=4$
$\text{Probability} = \frac{4}{16} = \frac{1}{4}$

Step5: Probability for sum<5

Favorable count = $1+2+3=6$
$\text{Probability} = \frac{6}{16} = \frac{3}{8}$

Answer:

1. $\frac{1}{8}$
2. $\frac{1}{4}$
3. $\frac{3}{8}$
4. $\frac{3}{16}$