Math Case Study for Class 8 Probability With Solutions

Math Case Study for Class 8 Probability With Solutions (Interactive Test)

Math Case Study for Class 8 Probability With Solutions

The Math Case Study for Class 8 Probability With Solutions helps students understand concepts through guided examples and solved case-based questions. Moreover, the chapter explains outcomes, events, and experiments in simple language. Some sentences stay short for clarity. Learners also connect probability to real-life situations, which strengthens understanding.

Step-by-Step Solved Examples

This section includes structured solutions that walk students through each method clearly. Additionally, step-by-step explanations make problem-solving easier. Many examples are designed to improve reasoning. Students gain confidence through repeated practice.

Practice Sheets and Answer Keys

The chapter contains MCQs, worksheets, and detailed answer keys. As a result, students can review their performance independently. Furthermore, regular practice improves accuracy. These resources make exam preparation simple and effective.

Math Case Study for Class 8 Probability With Solutions (Interactive Test)

The school organized a small mathematics fair where students learned **probability** through a simple prize game. Each student takes one turn. In a turn the student rolls one **fair six-sided die** (faces 1 to 6) and then flips one **fair coin** (Head or Tail). The rules are: if the die shows an **even number (2, 4, or 6)** the student wins a **cookie**; if the coin shows **Head** the student wins a **sticker**. If both happen, the student wins both cookie and sticker in that turn. The teacher asked the students to play many turns and to record the results. They discussed why each simple outcome in a single turn (a pair of die result and coin result) is equally likely, and how to use this fact to calculate theoretical probabilities. The class also compared the expected counts with their observed counts after many trials. They talked about how similar ideas appear in games, risk checks, and planning events where two or more independent things can happen together. This activity helped students practise using complements, multiplication rule for independent events, and expected value.

1. How many simple equally likely outcomes are there for one turn (one die roll and one coin flip)?

Solution:
Total outcomes = (Outcomes for Die) $\times$ (Outcomes for Coin) $= 6 \times 2 = 12$.
Correct answer is option **(c)**.

2. What is the probability that a student wins a cookie but not a sticker in one turn?

Solution:
Cookie: Die is Even ($\frac{3}{6} = \frac{1}{2}$). Not Sticker: Coin is Tail ($\frac{1}{2}$).
$P(\text{Cookie and Not Sticker}) = P(\text{Even}) \times P(\text{Tail}) = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$.
Correct answer is option **(b)**.

3. What is the probability that the student wins at least one prize (cookie or sticker or both) in one turn?

Solution:
Use the complement rule: $P(\text{at least one prize}) = 1 – P(\text{no prize})$.
No Prize: Die is Odd ($\frac{3}{6} = \frac{1}{2}$) AND Coin is Tail ($\frac{1}{2}$).
$P(\text{No Prize}) = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$.
$P(\text{At least one prize}) = 1 – \frac{1}{4} = \frac{3}{4}$.
Correct answer is option **(d)**.

4. If the class plays the game 300 times, how many turns with both cookie and sticker should they expect on average?

Solution:
$P(\text{Both Prizes}) = P(\text{Even on Die}) \times P(\text{Head}) = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$.
Expected number $= \text{Number of Trials} \times P(\text{Both Prizes}) = 300 \times \frac{1}{4} = 75$.
Correct answer is option **(c)**.

5. What is the probability of getting exactly one prize (either cookie only or sticker only) in one turn?

Solution:
$P(\text{Cookie Only}) = P(\text{Even and Tail}) = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$.
$P(\text{Sticker Only}) = P(\text{Odd and Head}) = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$.
$P(\text{Exactly One Prize}) = P(\text{Cookie Only}) + P(\text{Sticker Only}) = \frac{1}{4} + \frac{1}{4} = \frac{2}{4} = \frac{1}{2}$.
Correct answer is option **(b)**.

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