SOLUTION: Past records show that 4 of 135 parts are defective in length, 3 of 141 are defective in width, and 2 of 347 are defective in both. Use these figures
to estimate probabilities of
Algebra ->
Probability-and-statistics
-> SOLUTION: Past records show that 4 of 135 parts are defective in length, 3 of 141 are defective in width, and 2 of 347 are defective in both. Use these figures
to estimate probabilities of
Log On
Question 1179235: Past records show that 4 of 135 parts are defective in length, 3 of 141 are defective in width, and 2 of 347 are defective in both. Use these figures
to estimate probabilities of the individual events assuming that defects occur independently in length and width.
1) What are the fair odds against a defect (in length or width or both)?
You can put this solution on YOUR website! Let's break down this problem step-by-step:
**1. Calculate Individual Defect Probabilities:**
* **Defective in Length (L):**
* P(L) = 4 / 135 ≈ 0.0296
* **Defective in Width (W):**
* P(W) = 3 / 141 ≈ 0.0213
* **Defective in Both (L and W):**
* P(L and W) = 2 / 347 ≈ 0.0058
**2. Calculate the Probability of a Defect (L or W or Both):**
We'll use the formula for the probability of the union of two events:
P(L or W) = P(L) + P(W) - P(L and W)
P(L or W) ≈ 0.0296 + 0.0213 - 0.0058 ≈ 0.0451
**3. Calculate the Probability of No Defect:**
The probability of no defect is the complement of the probability of a defect:
P(No Defect) = 1 - P(L or W)
P(No Defect) ≈ 1 - 0.0451 ≈ 0.9549
**4. Calculate the Fair Odds Against a Defect:**
Fair odds are the ratio of the probability of the event not occurring to the probability of the event occurring:
Odds Against Defect = P(No Defect) / P(L or W)
Odds Against Defect ≈ 0.9549 / 0.0451 ≈ 21.17
To express this in a more standard odds format, we can approximate it as a ratio of whole numbers.
21.17/1 is close to 21/1.
**Answer:**
The fair odds against a defect (in length or width or both) are approximately 21 to 1.
