1. [3 points] A coin is flipped 70 times (the probabilities of heads and tails in each flip are the same). Let $p$ be the probability that heads will appear more than 42 times, and $q$ be the probability that heads will appear no more than 28 times. Find $p-q$.
Answer: $-\frac{1}{2^{70}} \cdot C_{70}^{28}$.

Solution. Since the probability of getting heads or tails is equal, the probability of getting 70 heads is equal to the probability of getting 70 tails (i.e., 0 heads); the probability of getting 69 heads is equal to the probability of getting 69 tails (i.e., one head), and so on. Let's denote the probability of getting exactly $k$ heads as $p_{k}$. Then $p=p_{43}+p_{44}+\ldots+p_{70}, q=p_{0}+p_{1}+\ldots+p_{28}$, and due to the above, $q=p_{70}+p_{69}+\ldots+p_{42}$. Therefore, $p-q=-p_{42}=-p_{28}$.

Let's calculate the probability that heads appear exactly 28 times in 70 tosses. If we denote the appearance of heads by 1 and the appearance of tails by 0, then each sequence of 70 tosses can be characterized by a sequence of digits consisting of zeros and ones. The probability of getting any such sequence is $\frac{1}{2^{70}}$. We are interested in all sequences of events that contain exactly 28 ones. The number of such sequences is $C_{70}^{28}$ (we choose 28 positions out of the available 70 for ones, and the remaining positions are filled with zeros). Therefore, the probability of getting at least one such sequence is $\frac{1}{2^{70}} \cdot C_{70}^{28}$. This is $p_{28}$.