On the exam, there are three trigonometry problems, two algebra problems, and five geometry problems. Vanya solves trigonometry problems with a probability of 

$p_{1}=0.2$, geometry problems with a probability of $p_{2}=0.4$, and algebra problems with a probability of $p_{3}=0.5$. To get a grade of three, 

Vanya needs to solve at least five problems.

a) What is the probability that Vanya will solve at least five problems?

Vanya decided to focus intensively on problems from one section. In a week, he can increase the probability of solving problems from this section by 0.2.

b) Which section should Vanya focus on to maximize the probability of solving at least five problems?

c) Which section should Vasya focus on to maximize the expected number of solved problems?
a) Let the random variable $X$ be the number of solved trigonometry problems, $Y$ be the number of solved geometry problems, and $Z$ be the number of solved algebra problems. Then the sum $U=X+Y+Z$ is the total number of solved problems. We want to find the probability $\mathrm{P}(U \geq 5)$. It is easier to compute

$\mathrm{P}(U<5)=\mathrm{P}(U=0)+\mathrm{P}(U=1)+\mathrm{P}(U=2)+\mathrm{P}(U=3)+\mathrm{P}(U=4)$.

Let's write the distribution tables for the random variables $X, Y$, and $Z$.

The probability of solving $0 \leq k \leq 3$ trigonometry problems is $C_{3}^{k} p_{1}^{3-k}\left(1-p_{1}\right)^{k}$.

| $X$ | 0 | 1 | 2 | 3 |
| :---: | :---: | :---: | :---: | :---: |
| $p_{X}$ | 0.512 | 0.384 | 0.096 | 0.008 |

The probability of solving $0 \leq n \leq 5$ geometry problems is $C_{5}^{n} p_{2}^{5-n}\left(1-p_{2}\right)^{n}$.

| $Y$ | 0 | 1 | 2 | 3 | 4 | 5 |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| $p_{Y}$ | 0.07776 | 0.2592 | 0.3456 | 0.2304 | 0.0768 | 0.01024 |

And the probability of solving $0 \leq m \leq 2$ algebra problems is $C_{2}^{m} p_{3}^{2-m}\left(1-p_{3}\right)^{m}$.

| $Z$ | 0 | 1 | 2 |
| :---: | :---: | :---: | :---: |
| $p_{Z}$ | 0.25 | 0.5 | 0.25 |

Now let's fill in a similar table for $U$, representing the number of solved problems as the sum of different terms. In this case,

$\mathrm{P}(X=k, Y=n, Z=m)=\mathrm{P}(X=k) \mathrm{P}(Y=n) \mathrm{P}(Z=m)$

| $U$ | $X$ | $Y$ | $Z$ | $\mathrm{P}(X=k ; Y=n ; Z=m)$ |
| :---: | :---: | :---: | :---: | :---: |
| 0 | 0 | 0 | 0 | 0.00995328 |
| 1 | 0 | 0 | 1 | 0.01990656 |
|  | 0 | 1 | 0 | 0.0331776 |
|  | 1 | 0 | 0 | 0.00746496 |
| 2 | 0 | 0 | 2 | 0.00995328 |
|  | 0 | 1 | 1 | 0.0663552 |
|  | 0 | 2 | 0 | 0.0442368 |
|  | 1 | 0 | 1 | 0.01492992 |
|  | 1 | 1 | 0 | 0.0248832 |
|  | 2 | 0 | 0 | 0.00186624 |
| 3 | 0 | 1 | 2 | 0.0331776 |
|  | 0 | 2 | 1 | 0.0884736 |
|  | 0 | 3 | 0 | 0.0294912 |
|  | 1 | 0 | 2 | 0.00746496 |
|  | 1 | 1 | 1 | 0.0497664 |
|  | 1 | 2 | 0 | 0.0331776 |
|  | 2 | 0 | 1 | 0.00373248 |
|  | 2 | 1 | 0 | 0.0062208 |
|  | 3 | 0 | 0 | 0.00015552 |
| 4 | 0 | 2 | 2 | 0.0442368 |
|  | 0 | 3 | 1 | 0.0589824 |
|  | 0 | 4 | 0 | 0.0098304 |
|  | 1 | 1 | 2 | 0.0248832 |
|  | 1 | 2 | 1 | 0.0663552 |
|  | 1 | 3 | 0 | 0.0221184 |
|  | 2 | 0 | 2 | 0.00186624 |
|  | 2 | 1 | 1 | 0.0124416 |
|  | 2 | 2 | 0 | 0.0082944 |
|  | 3 | 0 | 1 | 0.00031104 |
|  | 3 | 1 | 0 | 0.0005184 |

The probability $\mathrm{P}(U=l)$ of getting $l$ points is the sum of the probabilities of each case of partitioning $l$.

| $U$ | 0 | 1 | 2 | 3 | 4 |
| :---: | :---: | :---: | :---: | :---: | :---: |
| $p_{U}$ | 0.00131072 | 0.03678208 | 0.13899776 | 0.24180224 | 0.24983808 |

The desired probability is $\mathrm{P}(U \geq 5)=1-\mathrm{P}(U<5) \approx 0.27$.

b) In a way similar to that described in part a), we find that if the probability of solving trigonometry problems is increased by 0.2, the probability of solving at least 5 problems $\mathrm{P}(U \geq 5)$ will be $0.41752576 \approx 0.42$.

If the probability of solving geometry problems is increased by 0.2, we get $\mathrm{P}(U \geq 5)=0.52636928 \approx 0.53$.

![](https://cdn.mathpix.com/cropped/2024_05_08_22585010b2534c7ab9ebg-28.jpg?height=51&width=1896&top_left_y=2507&top_left_x=1)
Thus, Vanya should focus on geometry.

c) Let's compute the expected number of solved problems $\mathrm{E} U=\mathrm{E} X+\mathrm{E} Y+\mathrm{E} Z$. $\mathrm{E} X=\mathrm{E} X_{1}+\mathrm{E} X_{2}+\mathrm{E} X_{3}$, where $X_{i}$ is 1 with probability $p_{1}$ if the $i$-th trigonometry problem is solved, and 0 with probability $1-p_{1}$ if it is not solved. Thus, $\mathrm{E} X_{i}=p_{1}$, so $\mathrm{E} X=3 p_{1}$. Similarly, $\mathrm{E} Y=5 p_{2}, \mathrm{E} Z=2 p_{3}$, so $\mathrm{E} U=3 p_{1}+5 p_{2}+2 p_{3}$. Therefore, if $p_{1}$ is increased by 0.2, $\mathrm{E} U$ will increase by 0.6; if $p_{2}$ is increased by 0.2, $\mathrm{E} U$ will increase by 1; and if $p_{3}$ is increased by 0.2, $\mathrm{E} U$ will increase by 0.4.

## Answer

a) $\approx 0.27$; b)-c) geometry.

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