In Anchuria, a unified state exam is taking place. The probability of guessing the correct answer to each question on the exam is 0.25. In 2011, to obtain a certificate, one needed to answer three questions correctly out of 20. In 2012, the School Administration of Anchuria decided that three questions were too few. Now, one needs to answer six questions correctly out of 40. The question is, if one knows nothing and simply guesses the answers, in which year is the probability of obtaining an Anchurian certificate higher - in 2011 or in 2012?
If a graduate guesses the answers, the Unified State Exam (EGE) can be considered a Bernoulli scheme with a success probability of $p=0.25$ and a failure probability of $q=0.75$. In 2011, to pass the exam, one needed to answer at least three questions correctly. It is more convenient to find the probability of the opposite event, i.e., the probability of answering fewer than three questions correctly. This is the sum of the probabilities of answering 0, 1, and 2 questions correctly, i.e., $\quad \sum_{k=0}^{2} C_{20}^{k} p^{k} q^{20-k} \approx 0.091$; therefore, the probability of passing the exam is approximately 0.909. Similarly, in 2012, the probability of failing the exam is

$$
\sum_{k=0}^{5} C_{40}^{k} p^{k} q^{40-k} \approx 0.043, \text { and }
$$

the probability of passing the exam is approximately 0.957.

## Answer

![](https://cdn.mathpix.com/cropped/2024_05_06_309dcf72bd08bf7391b4g-13.jpg?height=220&width=2038&top_left_y=153&top_left_x=-1)

While waiting for customers, a watermelon seller sequentially weighed 20 watermelons (weighing 1 kg, 2 kg, 3 kg, ..., 20 kg), balancing the watermelon on one scale pan with one or two weights on the other pan (possibly identical). The seller recorded on a piece of paper the weights of the weights he used. What is the smallest number of different numbers that could have appeared in his records, if the weight of each weight is an integer number of kilograms?

## Solution

With one or two weights of 1 kg, 3 kg, 5 kg, 7 kg, 9 kg, and 10 kg, any of the given watermelons can be weighed. Indeed, $2=1+1, 4=3+1$, $6=5+1, 8=7+1, 11=10+1, 12=9+3, 13=10+3, 14=9+5, 15=10+5, 16=9+7, 17=10+7, 18=9+$ $9, 19=10+9, 20=10+10$. Thus, six different numbers could have been recorded.

We will show that five types of weights are insufficient for the required weighings. If there are five weights, then, generally speaking, twenty watermelons can be weighed. Specifically: five watermelons can be balanced with single weights, five with double weights, and the remaining $5 \cdot 4: 2=10$ watermelons with pairs of different weights. However, each combination of weights must be used exactly once.

Note that half of the watermelons have an odd mass. Let $k$ of the five weights have an odd mass, and $5-k$ have an even mass. Then the number of ways to weigh a watermelon with an odd mass is exactly $k+k(5-k)=6k-k^2$. However, for no $k=0,1,2,3,4,5$ does this expression equal 10 (this can be verified either by substitution or by solving the quadratic equation $\left.6k-k^2=10\right)$.

## Answer

6 numbers.