Math Problem Statement

Explain the Binomial Distribution with example. The probability that student knows the correct answer to a multiple choice question is 2/3. If the student does not know the answer, the student guesses the answer. The probability of guessed answer is correct is ¼ . Given that student has answered the question correctly. Calculate the conditional probability that student knows the correct answer.

Solution

Binomial Distribution Explanation

The Binomial Distribution is a discrete probability distribution that models the number of successes in a fixed number of independent Bernoulli trials, each with the same probability of success. A Bernoulli trial is an experiment where there are only two possible outcomes: success or failure.

The probability mass function (PMF) of the Binomial distribution is given by:

P(X=k)=(nk)pk(1p)nkP(X = k) = \binom{n}{k} p^k (1-p)^{n-k}

where:

  • XX is the random variable representing the number of successes,
  • nn is the number of trials,
  • kk is the number of successes,
  • pp is the probability of success on each trial,
  • (nk)\binom{n}{k} is the binomial coefficient, calculated as n!k!(nk)!\frac{n!}{k!(n-k)!}.

Example Scenario: Multiple Choice Question

Let's address the problem step by step.

Given:

  • Probability that the student knows the answer (P(K)P(K)) = 23\frac{2}{3}
  • Probability that the student guesses (P(G)P(G)) = 1P(K)=131 - P(K) = \frac{1}{3}
  • If the student knows the answer, the probability of a correct answer (P(CK)P(C \mid K)) = 1
  • If the student guesses, the probability of a correct answer (P(CG)P(C \mid G)) = 14\frac{1}{4}
  • The student has answered the question correctly (CC).

We need to find the conditional probability that the student knew the answer given that the student answered correctly, P(KC)P(K \mid C).

Bayes' Theorem

To solve this, we'll use Bayes' Theorem, which is stated as:

P(KC)=P(CK)P(K)P(C)P(K \mid C) = \frac{P(C \mid K) \cdot P(K)}{P(C)}

Where P(C)P(C) is the total probability of answering correctly. Using the law of total probability:

P(C)=P(CK)P(K)+P(CG)P(G)P(C) = P(C \mid K) \cdot P(K) + P(C \mid G) \cdot P(G)

Substitute the given values:

P(C)=123+1413P(C) = 1 \cdot \frac{2}{3} + \frac{1}{4} \cdot \frac{1}{3}

P(C)=23+112=812+112=912=34P(C) = \frac{2}{3} + \frac{1}{12} = \frac{8}{12} + \frac{1}{12} = \frac{9}{12} = \frac{3}{4}

Now, substitute back into Bayes' Theorem:

P(KC)=12334=2334=23×43=89P(K \mid C) = \frac{1 \cdot \frac{2}{3}}{\frac{3}{4}} = \frac{\frac{2}{3}}{\frac{3}{4}} = \frac{2}{3} \times \frac{4}{3} = \frac{8}{9}

Conclusion

The conditional probability that the student knew the correct answer given that they answered correctly is 89\frac{8}{9}.

Further Questions and Tip

  1. How is the Binomial Distribution used in real-world scenarios?
  2. What are the conditions required for a Binomial Distribution to be applicable?
  3. Can you explain how to calculate the mean and variance of a Binomial Distribution?
  4. How does the Binomial Distribution relate to the Normal Distribution?
  5. What is the significance of Bayes' Theorem in probability theory?

Tip: When solving problems involving conditional probabilities, always check if Bayes' Theorem applies—it’s a powerful tool for reversing conditional dependencies.

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Math Problem Analysis

Mathematical Concepts

Probability
Conditional Probability
Bayes' Theorem

Formulas

P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}
P(K \mid C) = \frac{P(C \mid K) \cdot P(K)}{P(C)}
P(C) = P(C \mid K) \cdot P(K) + P(C \mid G) \cdot P(G)

Theorems

Bayes' Theorem
Law of Total Probability

Suitable Grade Level

Grades 11-12