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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

| 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 as a Bernoulli scheme with the probability of success $p=$ 0.25 and the probability of failure $q=0.75$. 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, that is, the probability of answering fewer than three questions correctly. This probability is the sum of the probabilities that the graduate answered 0, 1, and 2 questions correctly, that is, $\quad \sum_{k=0}^{2} C_{20}^{k} p^{k} q^{20-k} \approx 0.091 ;$ therefore, the probability of passing the exam is approximately 0.909.

Similarly, we find that in 2012, the probability of failing the exam is

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

the probability of passing the exam is approximately $0.957$.

## Answer

In 2012.