# More On Expected Value - Grade 9 Statistics & Probability
Section A: Multiple Choice Questions
Q1: A fair six-sided die is rolled. What is the expected value of the outcome? (a) 3 (b) 3.5 (c) 4 (d) 6
Solution:
Ans: (b) Explanation: The expected value is calculated as \(E(X) = \sum x \cdot P(x)\). For a fair die: \(E(X) = 1(\frac{1}{6}) + 2(\frac{1}{6}) + 3(\frac{1}{6}) + 4(\frac{1}{6}) + 5(\frac{1}{6}) + 6(\frac{1}{6}) = \frac{21}{6} = 3.5\). Option (a) is the median, not the expected value. Option (c) is close but incorrect. Option (d) is the maximum value, not the expected value.
Q2: A lottery ticket costs $5. The probability of winning $100 is 0.02, and the probability of winning nothing is 0.98. What is the expected value of buying one ticket? (a) -$3.00 (b) $2.00 (c) -$5.00 (d) $0.00
Solution:
Ans: (a) Explanation: The expected value is calculated as: \(E(X) = (100)(0.02) + (0)(0.98) - 5 = 2 - 5 = -3\). The negative value indicates an expected loss of $3.00. Option (b) ignores the cost of the ticket. Option (c) assumes you always lose the ticket price. Option (d) incorrectly suggests breaking even.
Q3: If the expected value of a game is $0, the game is considered to be: (a) Unfavorable to the player (b) Fair (c) Favorable to the player (d) Invalid
Solution:
Ans: (b) Explanation: A game with an expected value of zero is called a fair game because, on average, the player neither gains nor loses money over time. Option (a) would require a negative expected value. Option (c) would require a positive expected value. Option (d) is incorrect as the game is valid.
Q4: A probability distribution has outcomes 2, 4, and 6 with probabilities 0.3, 0.5, and 0.2 respectively. What is the expected value? (a) 3.8 (b) 4.0 (c) 4.2 (d) 3.6
Solution:
Ans: (a) Explanation: Using the formula \(E(X) = \sum x \cdot P(x)\): \(E(X) = 2(0.3) + 4(0.5) + 6(0.2) = 0.6 + 2.0 + 1.2 = 3.8\). Option (b) is the middle value but not weighted by probabilities. Option (c) and (d) result from calculation errors.
Q5: If you multiply all outcomes of a probability distribution by a constant k, the new expected value is: (a) The same as the original expected value (b) k times the original expected value (c) k plus the original expected value (d) The original expected value divided by k
Solution:
Ans: (b) Explanation: This is the linearity property of expected value: \(E(kX) = kE(X)\). When each outcome is multiplied by a constant k, the expected value is also multiplied by k. Option (a) ignores the scaling effect. Option (c) confuses multiplication with addition. Option (d) reverses the operation.
Q6: A spinner has three sections: red (probability 0.4), blue (probability 0.35), and green (probability 0.25). You win $10 for red, $5 for blue, and lose $8 for green. What is the expected value? (a) $3.75 (b) $4.25 (c) $3.25 (d) $5.75
Solution:
Ans: (a) Explanation: Calculate: \(E(X) = 10(0.4) + 5(0.35) + (-8)(0.25) = 4 + 1.75 - 2 = 3.75\). The expected value is $3.75. Option (b) results from not properly handling the loss. Option (c) and (d) involve calculation errors.
Q7: If you add a constant c to all outcomes in a probability distribution, how does the expected value change? (a) It is multiplied by c (b) It is divided by c (c) It increases by c (d) It remains unchanged
Solution:
Ans: (c) Explanation: According to the linearity property: \(E(X + c) = E(X) + c\). Adding a constant to all outcomes increases the expected value by that same constant. Option (a) confuses addition with multiplication. Option (b) is incorrect. Option (d) ignores the shift in all values.
Q8: Two independent games have expected values of $4 and $7 respectively. What is the expected value if you play both games? (a) $28 (b) $5.50 (c) $11 (d) $3
Solution:
Ans: (c) Explanation: For independent events, the expected value of the sum equals the sum of the expected values: \(E(X + Y) = E(X) + E(Y) = 4 + 7 = 11\). Option (a) incorrectly multiplies the values. Option (b) calculates an average. Option (d) subtracts instead of adds.
Section B: Fill in the Blanks
Q9: The expected value is also known as the __________ of a probability distribution.
Solution:
Ans: mean Explanation: The expected value represents the long-run average or mean of a random variable's outcomes when weighted by their probabilities.
Q10: If all probabilities in a distribution are equal and there are n outcomes, each probability equals __________.
Solution:
Ans: \(\frac{1}{n}\) Explanation: In a uniform probability distribution, each of the n outcomes has equal probability, so each probability is \(\frac{1}{n}\), ensuring the sum of all probabilities equals 1.
Q11: The formula for expected value is \(E(X) = \sum x \cdot P(x)\), where x represents the __________ and P(x) represents the __________.
Solution:
Ans: outcome (or value), probability Explanation: In the expected value formula, x denotes each possible outcome or value of the random variable, and P(x) denotes the probability of that outcome occurring.
Q12: A game where the expected value is negative is considered __________ to the player.
Solution:
Ans: unfavorable Explanation: When the expected value is negative, the player can expect to lose money on average over time, making the game unfavorable or disadvantageous to the player.
Q13: For any probability distribution, the sum of all probabilities must equal __________.
Solution:
Ans: 1 Explanation: This is a fundamental property of probability distributions: \(\sum P(x) = 1\). All probabilities must account for 100% of possible outcomes.
Q14: The property \(E(aX + b) = aE(X) + b\) is called the __________ of expected value.
Solution:
Ans: linearity Explanation: The linearity property states that expected value is a linear operator, meaning constants can be factored out and added terms can be separated.
Section C: Word Problems
Q15: A basketball player has a 60% chance of making a free throw worth 1 point and a 40% chance of missing (0 points). If she attempts 50 free throws in a season, what is the expected total number of points she will score from free throws?
Solution:
Ans: First, find the expected value per free throw: \(E(X) = 1(0.60) + 0(0.40) = 0.60\) points For 50 attempts: \(50 \times 0.60 = 30\) points Final Answer: 30 points
Q16: A carnival game costs $3 to play. You spin a wheel with the following outcomes: win $15 with probability 0.1, win $5 with probability 0.2, and win $0 with probability 0.7. Should you play this game based on the expected value? Justify your answer.
Solution:
Ans: Calculate the expected value of winnings: \(E(\text{winnings}) = 15(0.1) + 5(0.2) + 0(0.7) = 1.5 + 1.0 + 0 = 2.5\) dollars Expected profit: \(2.5 - 3 = -0.5\) dollars Final Answer: No, you should not play because the expected value is -$0.50, meaning you expect to lose $0.50 per game on average.
Q17: An insurance company offers a policy for a $200 premium. If a claim occurs (probability 0.15), the company pays out $1,500. If no claim occurs (probability 0.85), the company pays nothing. What is the insurance company's expected profit per policy?
Solution:
Ans: Expected payout: \(E(\text{payout}) = 1500(0.15) + 0(0.85) = 225\) dollars Expected profit: \(200 - 225 = -25\) dollars Final Answer: -$25 (The company expects to lose $25 per policy on average, indicating this premium is too low.)
Q18: A game involves rolling two fair six-sided dice. You win an amount equal to the sum of the two dice in dollars. What is the expected value of your winnings?
Solution:
Ans: Each die has an expected value of 3.5. By linearity: \(E(X + Y) = E(X) + E(Y) = 3.5 + 3.5 = 7\) Final Answer: $7
Q19: A quiz has 5 multiple-choice questions, each with 4 options. Random guessing gives a 0.25 probability of getting each question correct (worth 2 points) and 0.75 probability of getting it wrong (0 points). What is the expected total score if a student guesses randomly on all questions?
Solution:
Ans: Expected value per question: \(E(X) = 2(0.25) + 0(0.75) = 0.5\) points For 5 questions: \(5 \times 0.5 = 2.5\) points Final Answer: 2.5 points
Q20: A stock investment has the following probability distribution for annual returns: 20% gain with probability 0.5, 5% gain with probability 0.3, and 10% loss with probability 0.2. If you invest $1,000, what is the expected value of your investment after one year?
Solution:
Ans: Expected return rate: \(E(R) = 0.20(0.5) + 0.05(0.3) + (-0.10)(0.2) = 0.10 + 0.015 - 0.02 = 0.095\) This is a 9.5% expected gain. Expected value: \(1000 + 1000(0.095) = 1000 + 95 = 1095\) dollars Final Answer: $1,095
Exam, Summary, Important questions, MCQs, shortcuts and tricks, past year papers, video lectures, Worksheet (with Solutions): More On Expected Value, Worksheet (with Solutions): More On Expected Value, Extra Questions, pdf , Sample Paper, practice quizzes, Objective type Questions, Previous Year Questions with Solutions, ppt, mock tests for examination, Semester Notes, Worksheet (with Solutions): More On Expected Value, Free, Viva Questions, study material;
