SOLUTION: A multiple choice test has 25 questions, each with 5 possible choices, exactly one of which is correct.
The test is marked by giving four points to each correct choice, and subtra
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Question 1162804: A multiple choice test has 25 questions, each with 5 possible choices, exactly one of which is correct.
The test is marked by giving four points to each correct choice, and subtracting one point for each
incorrect choice. No points are either given or subtracted for questions for which no choice is selected.
Mickey and Bianca each know the answer to 12 of the 25 questions, but are unsure about the rest.
Mickey leaves the other 13 unanswered, whereas Bianca answers them randomly, picking one choice
for each question with equal probabilities, independently between questions.
(a) Find the expectation and variance of the number of questions that Bianca gets correct.
(b) Find the expectation and variance of the mark that Bianca gets.
(c) Find the expectation and variance of the mark that Mickey gets.
Answer by CPhill(2264) (Show Source): You can put this solution on YOUR website!
To solve this problem, we treat Bianca's random guessing as a binomial distribution and Mickey's strategy as a constant.
### (a) Expectation and variance of Bianca's correct questions
For the 13 questions Bianca guesses, each has a probability $p = \frac{1}{5} = 0.2$ of being correct. Let $X$ be the number of questions she gets correct out of these $n = 13$ guesses. This follows a binomial distribution: $X \sim \text{Bin}(13, 0.2)$.
**Expectation:**
$$E[X] = n \cdot p = 13 \cdot 0.2 = 2.6$$
**Variance:**
$$\text{Var}(X) = n \cdot p \cdot (1 - p) = 13 \cdot 0.2 \cdot 0.8 = 2.08$$
Bianca also knows 12 questions are correct for certain. Let $C_B$ be her total correct answers: $C_B = 12 + X$.
* **$E[C_B] = 12 + 2.6 = 14.6$**
* **$\text{Var}(C_B) = \text{Var}(X) = 2.08$** (since the 12 known questions have 0 variance).
---
### (b) Expectation and variance of Bianca's mark
Let $M_B$ be Bianca's mark. For the 12 questions she knows, she earns $12 \times 4 = 48$ points. For the 13 she guesses, let $G_i$ be the mark for a single guess:
* $P(G_i = 4) = 0.2$
* $P(G_i = -1) = 0.8$
**Expectation of one guess ($E[G_i]$):**
$$E[G_i] = (4 \times 0.2) + (-1 \times 0.8) = 0.8 - 0.8 = 0$$
Since the expectation of a single guess is 0, the total expected mark for the 13 guesses is $13 \times 0 = 0$.
* **$E[M_B] = 48 + 0 = 48$**
**Variance of one guess ($\text{Var}(G_i)$):**
First, find $E[G_i^2]$:
$$E[G_i^2] = (4^2 \times 0.2) + ((-1)^2 \times 0.8) = (16 \times 0.2) + (1 \times 0.8) = 3.2 + 0.8 = 4$$
$$\text{Var}(G_i) = E[G_i^2] - (E[G_i])^2 = 4 - 0^2 = 4$$
Since the 13 guesses are independent, we sum the variances:
* **$\text{Var}(M_B) = 13 \times 4 = 52$**
---
### (c) Expectation and variance of Mickey's mark
Mickey only answers the 12 questions he knows. He leaves the other 13 blank, earning 0 points for them. His mark $M_M$ is a constant.
**Expectation:**
He gets exactly $12 \times 4$ points.
* **$E[M_M] = 48$**
**Variance:**
Since Mickey's mark is not subject to randomness (he knows the 12 are correct and the rest are definitely 0), there is no deviation from the mean.
* **$\text{Var}(M_M) = 0$**
---
### Summary Table
| Statistic | Bianca | Mickey |
| :--- | :--- | :--- |
| **Expected Mark** | 48 | 48 |
| **Variance of Mark** | 52 | 0 |
**Insight:** Both students have the same expected score because the penalty for guessing ($1/4$ of the points gained) perfectly offsets the $1/5$ chance of being right. However, Bianca's strategy is much "riskier" due to the high variance.
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