1. [3 points] A coin is flipped 90 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 55 times, and $q$ be the probability that heads will appear fewer than 35 times. Find $p-q$.
Answer: $\frac{1}{2^{90}} \cdot C_{90}^{35}$.

Solution. Since the probability of getting heads or tails is equal, the probability of getting 90 heads is equal to the probability of getting 90 tails (i.e., 0 heads); the probability of getting 89 heads is equal to the probability of getting 89 tails (i.e., one head), and so on. Let's denote the probability of getting exactly $k$ heads as $p_{k}$. Then $p=p_{55}+p_{56}+\ldots+p_{90}, q=p_{0}+p_{1}+\ldots+p_{34}$, and due to the above, $q=p_{90}+p_{89}+\ldots+p_{56}$. Therefore, $p-q=p_{55}$.

Let's calculate the probability that heads appear exactly 55 times in 90 tosses. If we denote the appearance of heads by 1 and the appearance of tails by 0, then each sequence of 90 tosses can be characterized by a sequence of digits consisting of zeros and ones. The probability of getting any such sequence is $\frac{1}{2^{90}}$. We are interested in all sequences that contain exactly 35 ones. The number of such sequences is $C_{90}^{35}$ (we choose 35 positions out of the available 90 for the ones, and the remaining positions are filled with zeros). Therefore, the probability of getting at least one such sequence is $\frac{1}{2^{90}} \cdot C_{90}^{35}$. This is $p_{55}$.