Question

a spinner has two equal sections, one green and one orange. the spinner is spun three times, resulting in the sample space s = {ggg, ggo, gog, ogg, goo, ogo, oog, ooo}. if the random variable x represents the number of times orange, o, is spun, which graph represents the probability distribution?

Explanation:

Step1: Determine possible values of \( X \)

The spinner is spun 3 times, so \( X \) (number of orange spins) can be 0, 1, 2, or 3.

Step2: Calculate probabilities for each \( X \)

• \( P(X = 0) \): All green (GGG). There's 1 such outcome. Total outcomes: \( 2^3 = 8 \). So \( P(X = 0) = \frac{1}{8} = 0.125 \).
• \( P(X = 1) \): One orange, two green. Outcomes: GGO, GOG, OGG (3 outcomes). So \( P(X = 1) = \frac{3}{8} = 0.375 \).
• \( P(X = 2) \): Two orange, one green. Outcomes: GOO, OGO, OOG (3 outcomes). So \( P(X = 2) = \frac{3}{8} = 0.375 \).
• \( P(X = 3) \): All orange (OOO). 1 outcome. So \( P(X = 3) = \frac{1}{8} = 0.125 \).

Step3: Match with the graphs

Check the heights (probabilities) for \( X = 0, 1, 2, 3 \). The first graph (top - left) has:

• \( X = 0 \): ~0.1 (close to 0.125)
• \( X = 1 \): ~0.2 (no, should be 0.375) – Wait, re - check. Wait, the third graph (middle - bottom) has:

Wait, the correct graph should have \( P(0)=0.125 \), \( P(1)=0.375 \), \( P(2)=0.375 \), \( P(3)=0.125 \). Looking at the graphs, the third graph (the one with \( P_x(x) \) axis from 0 to 0.40) has:

• \( X = 0 \): ~0.125 (height around 0.1 - 0.15)
• \( X = 1 \): ~0.375 (height around 0.35 - 0.40)
• \( X = 2 \): ~0.375 (height around 0.3 - 0.35)
• \( X = 3 \): ~0.125 (height around 0.1 - 0.15)

Wait, the first graph (top - left) has:

• \( X = 0 \): ~0.1
• \( X = 1 \): ~0.2
• \( X = 2 \): ~0.6
• \( X = 3 \): ~0.1. No, that's wrong.

Wait, let's re - calculate. Wait, the sample space \( S=\{GGG, GGO, GOG, OGG, GOO, OGO, OOG, OOO\} \), so 8 elements.

• \( X = 0 \): 1 outcome (GGG) → \( P = 1/8 = 0.125 \)
• \( X = 1 \): 3 outcomes (GGO, GOG, OGG) → \( P = 3/8 = 0.375 \)
• \( X = 2 \): 3 outcomes (GOO, OGO, OOG) → \( P = 3/8 = 0.375 \)
• \( X = 3 \): 1 outcome (OOO) → \( P = 1/8 = 0.125 \)

Now, looking at the graphs:

• Top - left graph: \( X = 0 \) height ~0.1, \( X = 1 \) ~0.2, \( X = 2 \) ~0.6, \( X = 3 \) ~0.1. Sum: 0.1 + 0.2 + 0.6 + 0.1 = 1, but probabilities for \( X = 1 \) and \( X = 2 \) are wrong.

• Top - right graph: Axes are wrong ( \( P_x(x) \) up to 3.5, which is not probability).

• Middle - bottom graph: \( P_x(x) \) from 0 to 0.40. \( X = 0 \): ~0.125, \( X = 1 \): ~0.375, \( X = 2 \): ~0.375, \( X = 3 \): ~0.125. This matches.

• Bottom - right graph: \( X = 1 \) height ~0.5, which is wrong.

Wait, the first graph (top - left) has \( X = 2 \) height ~0.6, which is 3/5 = 0.6, but our \( P(X = 2) = 0.375 \). Wait, maybe I misread the first graph. Wait, the first graph's \( P_x(x) \) axis: 0, 0.1, 0.2, 0.3, 0.4, 0.5, 0.6. \( X = 0 \): height ~0.1 (0.125 close), \( X = 1 \): ~0.2 (no, 0.375), \( X = 2 \): ~0.6 (no, 0.375), \( X = 3 \): ~0.1 (0.125 close). No.

Wait, the correct graph is the third one (the one with the \( P_x(x) \) scale from 0.00 to 0.40). Let's check the heights:

• \( X = 0 \): bar height ~0.125 (between 0.10 and 0.15)
• \( X = 1 \): bar height ~0.375 (close to 0.35 - 0.40)
• \( X = 2 \): bar height ~0.375 (close to 0.30 - 0.35)
• \( X = 3 \): bar height ~0.125 (between 0.10 and 0.15)

Yes, this matches our calculated probabilities.

Answer:

The third graph (the one with the \( P_x(x) \) axis ranging from 0.00 to 0.40, and \( X \) values 0, 1, 2, 3)