Question

what is the probability of a success and a failure for this experiment?$\bigcirc$ $p(\text{success})=\frac{1}{4}$; $p(\text{failure})=\frac{7}{8}LXB0\bigcirc$ $p(\text{success})=\frac{3}{4}$; $p(\text{failure})=\frac{1}{4}$$\bigcirc$ $p(\text{success})=\frac{7}{8}$; $p(\text{failure})=\frac{1}{8}$

Explanation:

Step1: Count total equal sections

Total sections = 8

Step2: Identify success sections

Assume labeled sections (4,5,6,7) are successes: 6 sections.
Wait, correction: Wait, no—wait, the unlabeled sections are 2, labeled are 6? No, wait: the circle has 8 equal parts. The labeled parts are 4,5,6,7 (4 sections), unlabeled are 4? No, no—wait, the question implies success is the labeled? No, wait, no: probability of success + failure =1. Wait, let's count: the circle has 8 equal slices. Let's count the slices: <point>171 172</point><point>175 82</point><point>245 77</point><point>350 82</point><point>345 177</point><point>270 200</point><point>190 212</point><point>145 102</point>—wait no, 8 slices. The labeled ones are 4,5,6,7: 4 slices? No, 4,5,6,7 are 4 slices, so 4 out of 8? No, wait the options have 7/8 and 1/8. Oh! Wait, the smallest slice is 1/8. The purple small slice is 1 slice, so if failure is that 1 slice, success is 7. Wait, no: let's check the options. The sum of success and failure must be 1.

Wait, let's calculate each option:

1. $\frac{1}{4}+\frac{7}{8}=\frac{2}{8}+\frac{7}{8}=\frac{9}{8}

eq1$ → invalid

1. $\frac{1}{4}+\frac{3}{4}=1$ → valid
2. $\frac{3}{4}+\frac{1}{4}=1$ → valid
3. $\frac{7}{8}+\frac{1}{8}=1$ → valid

Now, count the slices: the circle has 8 equal slices. The number of "success" slices (assuming the larger slices are success): 7 slices are large, 1 is tiny. So success is 7/8, failure 1/8. Wait, yes: the tiny purple slice is 1/8, others are 7 slices (each 1/8). So total success (7 slices) is $\frac{7}{8}$, failure (1 slice) is $\frac{1}{8}$.

Step3: Verify probability sum

$P(\text{success}) + P(\text{failure}) = \frac{7}{8}+\frac{1}{8}=1$, which is correct.

Answer:

P(success)=$\frac{7}{8}$; P(failure)=$\frac{1}{8}$ (the fourth option: P(success)=$\frac{7}{8}$; P(failure)=$\frac{1}{8}$)