Question

1. a game has two players. on each turn, a player spins two spinners to determine how far to move a game piece.

• each spinner is divided into 4 equal sections.
• a different integer from 1 to 4 is written on each section of one spinner.
• a different integer from -1 to 2 is written on each section of the other spinner.
• the player finds the sum of the integers shown and moves the game piece that number of spaces.
• the first player to land on or pass the \win\ space wins the game.

the figure shows the position of each player’s game pieces. player a needs to move 6 spaces or more to win and player b needs to move 3 spaces or more to win.
figure with boxes labeled a, b, win

Answer:

Since the problem is not fully presented (the question part after the setup is missing, like what is being asked about Player A and Player B, e.g., probability of winning, or who is more likely to win), we can't provide a complete solution. But if we assume a common question like "Which player has a greater probability of winning?" or "Find the probability that Player A wins" etc., here's how we would start:

Step 1: Determine the possible outcomes of the spinners

• Spinner 1 (let's call it \( S_1 \)): numbers \( \{1, 2, 3, 4\} \) (4 sections, each with a different integer from 1 - 4)
• Spinner 2 (let's call it \( S_2 \)): numbers \( \{-1, 0, 1, 2\} \) (wait, the problem says "a different integer from -1 to 2" – so integers from -1 to 2 are -1, 0, 1, 2? Wait, -1 to 2 inclusive has 4 integers: -1, 0, 1, 2. Yes, because \( 2 - (-1) + 1 = 4 \). So \( S_2 = \{-1, 0, 1, 2\} \)

The total number of possible outcomes when spinning both spinners is \( 4 \times 4 = 16 \) (by the fundamental counting principle, since each spinner has 4 independent outcomes).

Step 2: Find the sum for each pair of outcomes

We need to find all possible sums \( s = s_1 + s_2 \), where \( s_1 \in S_1 \) and \( s_2 \in S_2 \).

Let's list all possible sums:

• When \( s_1 = 1 \):
• \( s_2 = -1 \): \( 1 + (-1) = 0 \)
• \( s_2 = 0 \): \( 1 + 0 = 1 \)
• \( s_2 = 1 \): \( 1 + 1 = 2 \)
• \( s_2 = 2 \): \( 1 + 2 = 3 \)
• When \( s_1 = 2 \):
• \( s_2 = -1 \): \( 2 + (-1) = 1 \)
• \( s_2 = 0 \): \( 2 + 0 = 2 \)
• \( s_2 = 1 \): \( 2 + 1 = 3 \)
• \( s_2 = 2 \): \( 2 + 2 = 4 \)
• When \( s_1 = 3 \):
• \( s_2 = -1 \): \( 3 + (-1) = 2 \)
• \( s_2 = 0 \): \( 3 + 0 = 3 \)
• \( s_2 = 1 \): \( 3 + 1 = 4 \)
• \( s_2 = 2 \): \( 3 + 2 = 5 \)
• When \( s_1 = 4 \):
• \( s_2 = -1 \): \( 4 + (-1) = 3 \)
• \( s_2 = 0 \): \( 4 + 0 = 4 \)
• \( s_2 = 1 \): \( 4 + 1 = 5 \)
• \( s_2 = 2 \): \( 4 + 2 = 6 \)

Now let's list all sums: \( 0, 1, 2, 3, 1, 2, 3, 4, 2, 3, 4, 5, 3, 4, 5, 6 \)

Step 3: Determine the number of favorable outcomes for each player

• Player A needs to move 6 or more spaces. So we need sums \( \geq 6 \). Looking at the sums, the only sum \( \geq 6 \) is 6 (from \( s_1 = 4, s_2 = 2 \)). So number of favorable outcomes for A: 1.
• Player B needs to move 3 or more spaces. So we need sums \( \geq 3 \). Let's count the number of sums \( \geq 3 \):

Looking at the list of sums:

0, 1, 2, 3, 1, 2, 3, 4, 2, 3, 4, 5, 3, 4, 5, 6

Let's count each sum \( \geq 3 \):

• Sum = 3: Let's count how many times 3 appears. From the list:

When \( s_1=1, s_2=2 \): 3

\( s_1=2, s_2=1 \): 3; \( s_1=2, s_2=2 \): 4 (no, wait, \( s_1=2, s_2=1 \) is 3, \( s_1=2, s_2=2 \) is 4. Wait, let's re - list the sums with their counts:

Sum = 0: 1

Sum = 1: 2 (from \( s_1=1,s_2=0 \); \( s_1=2,s_2=-1 \))

Sum = 2: 3 (from \( s_1=1,s_2=1 \); \( s_1=2,s_2=0 \); \( s_1=3,s_2=-1 \))

Sum = 3: Let's see:

\( s_1=1,s_2=2 \): 3

\( s_1=2,s_2=1 \): 3

\( s_1=3,s_2=0 \): 3

\( s_1=4,s_2=-1 \): 3

So sum = 3: 4 times

Sum = 4:

\( s_1=2,s_2=2 \): 4

\( s_1=3,s_2=1 \): 4

\( s_1=4,s_2=0 \): 4

So sum = 4: 3 times

Sum = 5:

\( s_1=3,s_2=2 \): 5

\( s_1=4,s_2=1 \): 5

So sum = 5: 2 times

Sum = 6: 1 time

Now sum \( \geq 3 \): sum = 3 (4) + sum = 4 (3) + sum = 5 (2) + sum = 6 (1) = 4 + 3 + 2 + 1 = 10

Step 4: Calculate probabilities

• Probability that Player A wins: \( P(A) = \frac{\text{Number of favorable outcomes for A}}{\text{Total outcomes}} = \frac{1}{16} \)
• Probability that Player B wins: \( P(B) = \frac{\text{Number of favorable outcomes for B}}{\text{Total outcomes}} = \frac{10}{16} = \frac{5}{8} \)

If the question was "Which player is more likely to win?" or "Compare the probabilities", then Player B has a higher probability.

But since the original problem's question is not provided, we can only assume a common question and solve it as above. If the actual question is different (e.g., "Find the probability that Player A moves at least 6 spaces"), the answer would be \( \frac{1}{16} \), and for Player B moving at least 3 spaces, it's \( \frac{10}{16}=\frac{5}{8} \)

If you can provide the exact question (e.g., who has a better chance to win, find the probability for a specific player, etc.), we can give a more precise answer.