Math Problem Statement

Use the sample data in the table.

  Results from Experiments with Polygraph Instruments Did the Subject Actually Lie?   No (Did Not Lie)

Yes (Lied)

Positive test result (Polygraph test indicated that the subject lied.)

15 (false positive)

42 (true positive)

Negative test result  (Polygraph test indicated that the subject did not lie.)

32 (true negative)

9 (false negative)   If four of the test subjects are randomly selected without replacement, find the probability that, in each case, the polygraph indicated that the subject lied. 

Solution

To solve this problem, we need to find the probability that all four randomly selected subjects have a positive test result (indicating that they lied), given the provided data from the polygraph experiments.

Step 1: Understand the Data

From the table:

  • Positive test result (Polygraph indicated that the subject lied):

    • Did Not Lie (False Positive): 15
    • Lied (True Positive): 42
  • Negative test result (Polygraph indicated that the subject did not lie):

    • Did Not Lie (True Negative): 32
    • Lied (False Negative): 9

The total number of subjects with a positive test result (indicated that they lied) is: Positive test result=15+42=57\text{Positive test result} = 15 + 42 = 57

Step 2: Total Number of Subjects

The total number of test subjects is: Total number of subjects=15+42+32+9=98\text{Total number of subjects} = 15 + 42 + 32 + 9 = 98

Step 3: Calculate the Probability

We want to find the probability that all four selected subjects have a positive test result.

The probability that the first subject selected has a positive test result is: P(first positive)=5798P(\text{first positive}) = \frac{57}{98}

After selecting one subject with a positive result, there are 56 subjects with a positive result left, and 97 total subjects remaining.

The probability that the second subject also has a positive test result is: P(second positive)=5697P(\text{second positive}) = \frac{56}{97}

Similarly, for the third subject: P(third positive)=5596P(\text{third positive}) = \frac{55}{96}

And for the fourth subject: P(fourth positive)=5495P(\text{fourth positive}) = \frac{54}{95}

Step 4: Compute the Combined Probability

Since these selections are without replacement, we multiply the probabilities together:

P(all four positive)=5798×5697×5596×5495P(\text{all four positive}) = \frac{57}{98} \times \frac{56}{97} \times \frac{55}{96} \times \frac{54}{95}

Step 5: Calculate the Result

Let's compute the final probability:

P(all four positive)=57×56×55×5498×97×96×95P(\text{all four positive}) = \frac{57 \times 56 \times 55 \times 54}{98 \times 97 \times 96 \times 95}

Now, let's calculate this fraction:

P(all four positive)9,955,68088,259,0400.1128P(\text{all four positive}) \approx \frac{9,955,680}{88,259,040} \approx 0.1128

Conclusion

The probability that all four randomly selected subjects have a positive test result (indicating they lied) is approximately 0.1128 or 11.28%.

Would you like more details or have any questions?


Related Questions:

  1. What is the probability of selecting exactly two subjects who had a positive test result?
  2. How would the probability change if the selection was with replacement instead of without?
  3. What is the probability of selecting four subjects such that two are false positives and two are true positives?
  4. How do the probabilities change if the total number of subjects changes but the ratios stay the same?
  5. Can you calculate the probability of at least one positive test result in four selections?

Tip:

When calculating probabilities involving multiple selections without replacement, consider how the total number of possible outcomes changes with each selection.

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Math Problem Analysis

Mathematical Concepts

Probability
Combinatorics

Formulas

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Theorems

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Suitable Grade Level

Grades 10-12