Question 1162804
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).

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### (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:
* **$\text{Var}(M_B) = 13 \times 4 = 52$**

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### (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.
* **$\text{Var}(M_M) = 0$**

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### 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.