10. (9-11) 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) (1p) What is the probability that Vanya will solve at least five problems?

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

b) (2p) Which section should Vanya focus on to maximize the probability of solving at least five problems?

c) (2p) Which section should Vasya focus on to maximize the expected number of solved problems?
# Solution.

a) Let the random variable $X$ be the number of trigonometry problems solved, $Y$ be the number of geometry problems solved, and $Z$ be the number of algebra problems solved. Then the sum $U=X+Y+Z$ is the total number of problems solved. We want to find the probability $\mathrm{P}(U \geq 5)$. We will use the probability of the complementary event:

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

Then the desired probability is $\mathrm{P}(U \geq 5)=1-\mathrm{P}(U<5)$.

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

$$
C_{2}^{m} p_{1}^{2-m}\left(1-p_{1}\right)^{m}
$$

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

Now we will fill in a similar table for $U$, representing the total number of problems solved as different sums. 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 of getting $l$ points $\mathrm{P}(U=l)$ is the sum of the probabilities of each case of partitioning $l$.

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

$$
\begin{gathered}
\mathrm{P}(U<5)=0.00131072+0.03678208+0.13899776+ \\
\quad+0.24180224+0.24983808=0.73422528
\end{gathered}
$$

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

Answer: $\approx 0.27$.

b) Using a method similar to the one described in part a), we find that if the probability of solving trigonometry problems $p_{1}$ 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 $p_{2}$ is increased by 0.2, we get $\mathrm{P}(U \geq 5)=0.52636928 \approx 0.53$.

If the probability of solving algebra problems $p_{3}$ is increased by 0.2, we get $\mathrm{P}(U \geq 5)=0.35814546 \approx 0.36$. Therefore, Vanya should focus on geometry.

Answer: geometry.

c) Let's calculate the expected number of problems solved

$$
\begin{gathered}
\mathrm{M} U=\mathrm{M} X+\mathrm{M} Y+\mathrm{M} Z \\
\mathrm{M} X=\mathrm{M} X_{1}+\mathrm{M} X_{2}+\mathrm{M} X_{3}
\end{gathered}
$$

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, $\mathrm{M} X_{i}=0 \cdot\left(1-p_{1}\right)+1 \cdot p_{1}=p_{1}$, so $\mathrm{M} X=3 p_{1}$. Similarly, $\mathrm{M} Y=5 p_{2}$ and $\mathrm{M} Z=2 p_{3}$.

If $p_{1}$ is increased by 0.2, we get $\mathrm{M} U=3\left(p_{1}+0.2\right)+5 p_{2}+2 p_{3}=4.2$.

If $p_{2}$ is increased by 0.2, we get $\mathrm{M} U=3 p_{1}+5\left(p_{2}+0.2\right)+2 p_{3}=4.6$.

If $p_{3}$ is increased by 0.2, we get $\mathrm{M} U=3 p_{1}+5 p_{2}+2\left(p_{3}+0.2\right)=4$.

Thus, the expected number of problems solved will be the highest if Vanya focuses on geometry.

Answer: geometry.