Calculating the Probability of Passing a True/False Test by Guessing

Often, students may be faced with the challenge of answering a set of true/false questions. The stakes can be high if a student is required to pass the test by guessing on each question. In this article, we explore the statistical probability involved in passing such a test.

Understanding the Problem

A test consists of 10 true/false questions. To pass the test, a student must answer at least 6 questions correctly. If a student guesses on each question, what is the probability that the student will pass the test?

Modeling the Problem with a Binomial Distribution

To solve this problem, we can use the binomial distribution. The binomial distribution is used to find the probability of a specific number of successes in a fixed number of independent trials. In this case:

Number of trials n 10 (number of questions) Probability of success p 0.5 (since there are two choices: true or false) Probability of failure q 1 - p 0.5

The binomial probability mass function is defined as:

P(X k) binom{n}{k} p^k q^{n-k}

Where binom{n}{k} is the binomial coefficient, representing the number of ways to choose k successes in n trials.

Calculating the Probability of Passing the Test

We want to calculate the probability of getting at least 6 questions correct. This means we need to calculate the probability of getting 6, 7, 8, 9, or 10 questions right. The formula for each case is as follows:

P(X ge; 6) P(X 6) P(X 7) P(X 8) P(X 9) P(X 10)

For k 6

P(X 6) binom{10}{6} (0.5)^6 (0.5)^4 binom{10}{6} (0.5)^{10}

binom{10}{6} 210 rarr; P(X 6) 210 times frac{1}{1024} frac{210}{1024}

For k 7

P(X 7) binom{10}{7} (0.5)^7 (0.5)^3 binom{10}{7} (0.5)^{10}

binom{10}{7} 120 rarr; P(X 7) 120 times frac{1}{1024} frac{120}{1024}

For k 8

P(X 8) binom{10}{8} (0.5)^8 (0.5)^2 binom{10}{8} (0.5)^{10}

binom{10}{8} 45 rarr; P(X 8) 45 times frac{1}{1024} frac{45}{1024}

For k 9

P(X 9) binom{10}{9} (0.5)^9 (0.5)^1 binom{10}{9} (0.5)^{10}

binom{10}{9} 10 rarr; P(X 9) 10 times frac{1}{1024} frac{10}{1024}

For k 10

P(X 10) binom{10}{10} (0.5)^{10} (0.5)^0 binom{10}{10} (0.5)^{10}

binom{10}{10} 1 rarr; P(X 10) 1 times frac{1}{1024} frac{1}{1024}

Summing the Probabilities

The probability of passing the test is the sum of the probabilities of getting 6, 7, 8, 9, or 10 questions correct:

P(X ge; 6) frac{210}{1024} frac{120}{1024} frac{45}{1024} frac{10}{1024} frac{1}{1024}

P(X ge; 6) frac{386}{1024}

P(X ge; 6) frac{386}{1024} frac{193}{512} approx 0.377

Conclusion

The probability that a student will pass the test by guessing is approximately 0.377, or 37.7%. This means that if a student guesses on each question, they have just under a 38% chance of passing the test.