1. [3 points] A coin is flipped 110 times (the probabilities of heads and tails in each flip are the same). Let $p$ be the probability that heads will appear no fewer than 61 times, and $q$ be the probability that heads will appear fewer than 49 times. Find $p-q$.
Answer: $\frac{1}{2^{110}} \cdot C_{100}^{61}$.

Solution. Since the probability of getting heads or tails is equal, the probability of getting 110 heads is equal to the probability of getting 110 tails (i.e., 0 heads); the probability of getting 109 heads is equal to the probability of getting 109 tails (i.e., one head), and so on. Let's denote the probability of getting exactly $k$ heads as $p_{k}$. Then $p=p_{61}+p_{62}+\ldots+p_{110}, q=p_{0}+p_{1}+\ldots+p_{48}$, and due to the above, $q=p_{100}+p_{109}+\ldots+p_{62}$. Therefore, $p-q=p_{61}$.

Let's calculate the probability that heads appear exactly 61 times in 110 tosses. If we denote the appearance of heads by 1 and the appearance of tails by 0, then each sequence of 110 tosses can be characterized by a sequence of digits consisting of zeros and ones. The probability of getting any such sequence is $\frac{1}{2^{110}}$. We are interested in all sequences of events that contain exactly 61 ones. The number of such sequences is $C_{110}^{61}$ (we choose 61 positions out of the available 110 for ones, and the remaining positions are filled with zeros). Therefore, the probability of getting at least one such sequence is $\frac{1}{2^{110}} \cdot C_{100}^{61}$. This is $p_{61}$.