Question

c) p(even or a multiple of
d) p(less than 10 and odd)
a) p(odd and unshaded)
b) p(shaded and at least
c) p(multiple of 6))

Explanation:

To solve these probability problems, we assume we are dealing with a set of numbers (likely from 1 to 20 or a similar range, given the denominators like 20 in part d). Let's solve part (c) \( P(\text{multiple of 6}) \) as an example (assuming the total number of outcomes \( n(S) = 20 \), common in such problems):

Step 1: Identify multiples of 6

Multiples of 6 within a reasonable range (e.g., 1–20) are: 6, 12, 18. So there are 3 favorable outcomes.

Step 2: Calculate probability

Probability \( P(A) = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} \).
If \( n(S) = 20 \), then:
\( P(\text{multiple of 6}) = \frac{3}{20} = 0.15 \) (or 15%).

Note for other parts:

• For part (a) \( P(\text{odd and unshaded}) \), we need to know which numbers are odd and unshaded (missing diagram details, but the process is: count odd + unshaded, divide by total).
• For part (b) \( P(\text{shaded and at least [X]}) \), define "at least X" (e.g., at least 10) and count shaded numbers meeting that, divide by total.
• For part (d) \( P(\text{less than 10 and odd}) \), numbers less than 10 and odd: 1, 3, 5, 7, 9 (5 numbers). If total is 20, \( P = \frac{5}{20} = \frac{1}{4} \) (matches the handwritten note).

Example Answer (for part c, assuming total = 20):

\( \boldsymbol{\frac{3}{20}} \) (or 0.15, 15%).

If the total number of outcomes differs (e.g., 1–12), adjust accordingly. For example, in 1–12, multiples of 6 are 6, 12 (2 numbers), so \( P = \frac{2}{12} = \frac{1}{6} \approx 0.167 \).

Please provide the full context (e.g., range of numbers, shaded/unshaded details) for precise answers to all parts.

Answer:

To solve these probability problems, we assume we are dealing with a set of numbers (likely from 1 to 20 or a similar range, given the denominators like 20 in part d). Let's solve part (c) \( P(\text{multiple of 6}) \) as an example (assuming the total number of outcomes \( n(S) = 20 \), common in such problems):

Step 1: Identify multiples of 6

Multiples of 6 within a reasonable range (e.g., 1–20) are: 6, 12, 18. So there are 3 favorable outcomes.

Step 2: Calculate probability

Probability \( P(A) = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} \).
If \( n(S) = 20 \), then:
\( P(\text{multiple of 6}) = \frac{3}{20} = 0.15 \) (or 15%).

Note for other parts:

• For part (a) \( P(\text{odd and unshaded}) \), we need to know which numbers are odd and unshaded (missing diagram details, but the process is: count odd + unshaded, divide by total).
• For part (b) \( P(\text{shaded and at least [X]}) \), define "at least X" (e.g., at least 10) and count shaded numbers meeting that, divide by total.
• For part (d) \( P(\text{less than 10 and odd}) \), numbers less than 10 and odd: 1, 3, 5, 7, 9 (5 numbers). If total is 20, \( P = \frac{5}{20} = \frac{1}{4} \) (matches the handwritten note).

Example Answer (for part c, assuming total = 20):

\( \boldsymbol{\frac{3}{20}} \) (or 0.15, 15%).

If the total number of outcomes differs (e.g., 1–12), adjust accordingly. For example, in 1–12, multiples of 6 are 6, 12 (2 numbers), so \( P = \frac{2}{12} = \frac{1}{6} \approx 0.167 \).

Please provide the full context (e.g., range of numbers, shaded/unshaded details) for precise answers to all parts.