Probability Case Study questions Class 12 PDF

Probability Case Study Questions for Class 12 – Printable PDF Format

Probability Case Study Questions Class 12 PDF

Probability Case Study questions Class 12 PDF helps students understand practical applications of probability concepts. Moreover, these questions enhance reasoning skills and exam confidence. They are designed as per the latest CBSE competency-based pattern. Therefore, students gain clarity in problem-solving. Short calculations improve speed.

Concepts Covered in Probability Case Studies

Case study questions include events, outcomes, and conditional probability. For example, problems may involve dice, cards, or data interpretation. As a result, students apply formulas logically. In addition, real-life situations make learning engaging. Thus, conceptual understanding improves steadily.

Preparation Strategy for Better Results

Firstly, revise important probability formulas regularly. Then, practice solved examples step by step. Meanwhile, focus on understanding the given data. Additionally, reviewing similar questions improves accuracy. Consequently, students perform better in board examinations.

Case Study 2: Probability Case Study questions Class 12 PDF

Case Study Description: A public health department is evaluating a new, non-invasive diagnostic test for a very rare genetic disorder, let’s call it Disease $D$. The prevalence of Disease $D$ in the general population is estimated to be $0.5\%$ (or one in two hundred people). The test is known to be highly accurate. Specifically, the probability that the test correctly identifies a person \textbf{with} the disease (a true positive result) is $99\%$. However, no test is perfect. The probability that the test correctly identifies a person \textbf{without} the disease (a true negative result) is $98\%$. This means there is a chance of a false positive (testing positive when healthy) and a false negative (testing negative when sick). A person is selected at random from the population and is administered this test. The public health officials are particularly interested in the probability that a person who tests positive actually has the disease, as this impacts resource allocation and patient counseling. This scenario highlights how \textbf{Conditional Probability} and the concept of \textbf{Bayes’ Theorem} are crucial for interpreting medical test results, especially for rare diseases, where the prior probability is very low. Even a highly accurate test can yield surprising results when applied to the general population. Let $D$ be the event that a person has the disease, and $T$ be the event that the test result is positive. We are given $P(D)=0.005$, $P(T|D)=0.99$, and $P(T’|D’)=0.98$.

Theory and Formulae Related to Conditional Probability:

  • **Prior Probability (Prevalence)**: $P(D)$
  • **Complementary Events**: $P(D’) = 1 – P(D)$ and $P(T) = 1 – P(T’)$
  • **Multiplication Theorem (Joint Probability)**: $P(D \cap T) = P(D) \cdot P(T|D)$
  • **Theorem of Total Probability**: $P(T) = P(D)P(T|D) + P(D’)P(T|D’)$
  • **Bayes’ Theorem (Posterior Probability)**: $P(D|T) = \frac{P(D \cap T)}{P(T)}$

1. What is the probability of a false positive result, $P(T|D’)$, i.e., testing positive given they do not have the disease?

Solution: The false positive rate $P(T|D’)$ is complementary to the true negative rate $P(T’|D’)$. $$P(T|D’) = 1 – P(T’|D’) = 1 – 0.98 = 0.02$$ Correct answer is option **(c)**.

2. What is the probability that a randomly selected person is disease-free and tests positive? $P(D’ \cap T)$

Solution: Using Multiplication Theorem: $P(D’) = 1 – P(D) = 1 – 0.005 = 0.995$. $$P(D’ \cap T) = P(D’) \cdot P(T|D’) = 0.995 \times 0.02 = 0.01990$$ Correct answer is option **(c)**.

3. What is the probability that a randomly selected person from the population tests positive for the disease? $P(T)$ (Total Probability)

Solution: Using the Theorem of Total Probability: $$P(T) = P(D)P(T|D) + P(D’)P(T|D’)$$ True Positive: $P(D \cap T) = 0.005 \times 0.99 = 0.00495$ False Positive: $P(D’ \cap T) = 0.01990$ (from Q2) $$P(T) = 0.00495 + 0.01990 = 0.02485$$ Correct answer is option **(b)**.

4. If a person tests positive, what is the probability that they actually have the disease? $P(D|T)$ (Bayes’ Theorem)

Solution: Using Bayes’ Theorem: $$P(D|T) = \frac{P(D \cap T)}{P(T)} = \frac{0.00495}{0.02485}$$ Convert to fractions and simplify: $$\frac{495}{2485} = \frac{99}{497}$$ Correct answer is option **(a)**.

5. Suppose 10 people are randomly selected and tested. Let $X$ be the random variable representing the number of people who test positive. What is the expected number (Mean) of people who will test positive?

Solution: $X$ follows a Binomial distribution $B(n, p)$, where $n=10$ and $p = P(\text{Test Positive}) = P(T) = 0.02485$ (from Q3). The Expected Value (Mean) is $E(X) = n \cdot p$. $$E(X) = 10 \times 0.02485$$ Correct answer is option **(c)**.

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