Question

in one round of an experiment, a spinner is spun, a coin is tossed, and a number pyramid is thrown. - the spinner has 3 equal sections labeled a, b, c. - the coin has heads and tails. - the number pyramid has equal sides labeled 1 through 4. part a how many possible outcomes are there? enter the answer in the box. part b how many possible ways can the outcome be a vowel and either tails or an odd number? enter the answer in the box.

Explanation:

Part A

Step1: Determine outcomes for each event

• Spinner: 3 outcomes (A, B, C)
• Coin: 2 outcomes (heads, tails)
• Number pyramid: 4 outcomes (1, 2, 3, 4)

Step2: Apply multiplication principle

Total outcomes = \( 3 \times 2 \times 4 \)
\( 3 \times 2 = 6 \), \( 6 \times 4 = 24 \)

Step1: Identify vowel on spinner

Vowel on spinner: only A (since A, B, C; A is vowel)

Step2: Analyze coin and pyramid for "tails or odd number"

• Coin: tails (1 outcome)
• Pyramid: odd numbers (1, 3) → 2 outcomes
• "Tails or odd number" includes:
• Tails (any number: 4 outcomes)
• Odd number (any coin: 2 outcomes, but overlap with tails & odd: 2 outcomes)
• Using inclusion - exclusion: \( 4 + 2 - 2 = 4 \)? Wait, better:
• Outcomes for (coin = tails) OR (pyramid = odd):
• Tails + any number: 4
• Heads + odd number: 2
• Total: \( 4 + 2 = 6 \)

Step3: Calculate total for vowel (A) and (tails or odd)

Vowel (A: 1 outcome) × (tails or odd: 6 outcomes) = \( 1 \times 6 = 6 \)? Wait, no:
Wait, correct approach:

• Vowel on spinner: A (1 option)
• For the coin and pyramid, we want (tails) OR (odd number). Let's list all possibilities:
• Coin = tails:
• Pyramid: 1, 2, 3, 4 (4 outcomes)
• Coin = heads:
• Pyramid: 1, 3 (2 outcomes, since odd)
• Total for (tails or odd): \( 4 + 2 = 6 \)
• Now, spinner is A (1 outcome), so total ways: \( 1 \times 6 = 6 \)? Wait, no, wait:

Wait, another way:

• Spinner: A (1)
• Coin: 2 options (heads, tails)
• Pyramid: 4 options (1 - 4)

We want (spinner = A) AND [(coin = tails) OR (pyramid is odd)]
Using logic: \( P(A) \times [P(tails) + P(odd) - P(tails \cap odd)] \)

• \( P(tails) = \frac{1}{2} \), \( P(odd) = \frac{2}{4}=\frac{1}{2} \), \( P(tails \cap odd)=\frac{1}{2} \times \frac{2}{4}=\frac{1}{4} \)

But in terms of counts:

• Spinner: 1
• Coin - pyramid combinations for (tails or odd):
• Tails (any pyramid): 4
• Heads (odd pyramid): 2
• Total: 4 + 2 = 6 (since tails and odd: when tails and odd, it's included in tails, so no double - count if we do tails (4) + heads (odd: 2))
• So total: 1 (spinner) × 6 (coin - pyramid) = 6? Wait, no, wait:

Wait, let's list all possible triples (spinner, coin, pyramid) where spinner = A and (coin = tails or pyramid is odd):

• Spinner = A, Coin = tails, Pyramid = 1
• Spinner = A, Coin = tails, Pyramid = 2
• Spinner = A, Coin = tails, Pyramid = 3
• Spinner = A, Coin = tails, Pyramid = 4
• Spinner = A, Coin = heads, Pyramid = 1
• Spinner = A, Coin = heads, Pyramid = 3

That's 6 outcomes. Wait, but let's check:

• Tails + any pyramid (4) + heads + odd (2) = 6. And spinner is A (1), so 1×6 = 6.

Wait, but wait, is that correct?
Wait, another way:

• Spinner: A (1)
• Coin: 2 choices, Pyramid: 4 choices.

We want (coin = tails) OR (pyramid ∈ {1, 3})
The number of (coin, pyramid) pairs where (coin = tails) OR (pyramid ∈ {1, 3}):

• Let S be the set of (coin, pyramid) pairs. |S| = 2×4 = 8
• Let T be (coin = tails): |T| = 4
• Let O be (pyramid ∈ {1, 3}): |O| = 2×2 = 4? Wait, no: pyramid has 4 sides, so for pyramid ∈ {1, 3}, coin can be heads or tails, so |O| = 2 (coin) × 2 (pyramid) = 4
• T ∩ O: (coin = tails) AND (pyramid ∈ {1, 3}): |T ∩ O| = 2
• By inclusion - exclusion: |T ∪ O| = |T| + |O| - |T ∩ O| = 4 + 4 - 2 = 6
• So (spinner = A) × |T ∪ O| = 1×6 = 6

Yes, so total ways: 6

Answer:

24

Part B