Probability case study questions Class 12 with solutions

Class 12 Probability Case Study Questions with Answers for Exam Practice

Probability Case Study Questions Class 12 with Solutions

Probability case study questions Class 12 with solutions help students understand real-life applications of probability concepts. Moreover, these questions enhance logical thinking and exam readiness. They follow the latest CBSE competency-based format. Therefore, students can apply formulas accurately. Short questions improve confidence.

Key Concepts Covered in Case Studies

Case study questions focus on events, outcomes, and conditional probability. For example, scenarios may include dice, cards, or data interpretation. As a result, students learn practical problem-solving. In addition, such questions strengthen conceptual clarity. Thus, understanding improves gradually.

Tips to Score Better in Probability

Firstly, revise important probability formulas thoroughly. Then, practice solved examples step by step. Meanwhile, analyze the given data carefully. Additionally, solving similar questions boosts accuracy. Consequently, students perform better in Class 12 board exams.

Case Study 3: Probability case study questions Class 12 with solutions

Case Study Description: A large e-commerce platform monitors customer activity and satisfaction. They know that **$70\%$ of all customers who visit their website place an order**. Of the customers who place an order, **$90\%$ express high satisfaction** with the delivery and product quality. However, of the customers who **do not** place an order, only **$20\%$ express high satisfaction** (perhaps due to finding what they needed elsewhere or enjoying the browsing experience).

The platform is interested in understanding the overall satisfaction rate and the relationship between placing an order and expressing high satisfaction. Furthermore, they decide to track a group of 5 customers who visit the site and analyze the probability distribution of those who express high satisfaction. This scenario is crucial for applying concepts like **conditional probability**, checking for **independent events**, and calculating the **mean** and **variance of a random variable** within a practical business context. The company must determine if satisfaction is significantly influenced by whether a purchase was made. If the events are not independent, placing an order is a strong predictor of satisfaction.

Let $O$ be the event that a customer places an order ($P(O) = 0.70$), and $S$ be the event that a customer expresses high satisfaction. Given: $P(S|O) = 0.90$, and $P(S|O’) = 0.20$.

Theory and Formulae Related to Probability and Random Variables:

  • **Multiplication Theorem (Joint Probability)**: $P(A \cap B) = P(A) \cdot P(B|A)$
  • **Theorem of Total Probability**: $P(S) = P(O)P(S|O) + P(O’)P(S|O’)$
  • **Independence Check**: Events $O$ and $S$ are independent if $P(S|O) = P(S)$.
  • **Binomial Distribution Mean**: $E(X) = n \cdot p$
  • **Binomial Distribution Variance**: $\text{Var}(X) = n \cdot p \cdot q$ (where $q = 1-p$)

1. What is the probability that a customer does not place an order and also expresses high satisfaction? $P(O’ \cap S)$

Solution: Using Multiplication Theorem: $P(O’) = 1 – 0.70 = 0.30$. $$P(O’ \cap S) = P(O’) \cdot P(S|O’) = 0.30 \times 0.20 = 0.06$$ Correct answer is option **(b)**.

2. What is the overall probability that a randomly selected customer expresses high satisfaction ($P(S)$)? (Total Probability)

Solution: Using the Theorem of Total Probability: $$P(S) = P(O)P(S|O) + P(O’)P(S|O’)$$ With Order: $0.70 \times 0.90 = 0.63$ Without Order: $0.30 \times 0.20 = 0.06$ (from Q1) $$P(S) = 0.63 + 0.06 = 0.69$$ Correct answer is option **(a)**.

3. Are the events $O$ (Placing an order) and $S$ (High satisfaction) independent?

Solution: Events are independent if $P(S|O) = P(S)$. We have $P(S|O) = 0.90$ and $P(S) = 0.69$. Since $0.90 \neq 0.69$, the events are **not independent**. The correct reason for non-independence in the options is **(c)**. Correct answer is option **(c)**.

4. A customer expresses high satisfaction. What is the probability that they did not place an order? $P(O’|S)$ (Bayes’ Theorem)

Solution: Using Bayes’ Theorem: $$P(O’|S) = \frac{P(O’ \cap S)}{P(S)}$$ From Q1: $P(O’ \cap S) = 0.06$. From Q2: $P(S) = 0.69$. $$P(O’|S) = \frac{0.06}{0.69} = \frac{6}{69}$$ Correct answer is option **(a)**.

5. For the group of 5 customers (independent trials), let $X$ be the number of customers who express high satisfaction. What is the variance of $X$? (Random Variable and Variance)

Solution: $X$ follows a Binomial distribution $B(n, p)$ with $n=5$ and success probability $p = P(S) = 0.69$. Probability of failure is $q = 1 – p = 1 – 0.69 = 0.31$. The Variance is given by: $$\text{Var}(X) = n \cdot p \cdot q = 5 \times 0.69 \times 0.31$$ Correct answer is option **(a)**.

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