1. **Problem 2.3:** Find the probability that, when selecting 3 balls from a box containing $a$ white, $b$ black, and $c$ red balls, at least two balls are of the same color. 2. Total number of balls is $a+b+c$. Total ways to choose 3 balls: $$\binom{a+b+c}{3}.$$ 3. To find the probability of at least two balls of the same color, use complementary probability: find probability of all 3 balls being of different colors and subtract from 1. 4. Number of ways to choose 3 balls all different colors (assuming $a,b,c \geq 1$): $$\binom{a}{1} \cdot \binom{b}{1} \cdot \binom{c}{1} = abc.$$ 5. Probability of all different colors: $$\frac{abc}{\binom{a+b+c}{3}}.$$ 6. Hence, probability of at least two balls the same color is: $$1 - \frac{abc}{\binom{a+b+c}{3}}.$$ --- 1. **Problem 2.4:** Two marksmen each fire 2 shots. Let $p$ be the probability a marksman hits the target each shot. Find probabilities $P_1$ and $P_2$ for each marksman hitting at least once. 2. Probability a marksman misses a shot is $1-p$. Probability misses both shots: $(1-p)^2.$ 3. Probability hits at least once (complement): $$P = 1 - (1-p)^2 = 2p - p^2.$$ 4. Thus, $P_1 = P_2 = 2p - p^2.$ --- 1. **Problem 2.5:** From a box with 15 white and 10 black balls, selecting 8 balls, find probability that white balls in selection exceed black balls. 2. Total balls $= 25$, total ways: $$\binom{25}{8}.$$ 3. Let $x$ be white balls chosen, so black balls chosen are $8 - x$. We want $x > 8 - x \Rightarrow x > 4$. So $x = 5,6,7,8$. 4. Probability: $$\frac{\sum_{x=5}^8 \binom{15}{x} \binom{10}{8 - x}}{\binom{25}{8}}.$$ --- 1. **Problem 2.6:** From a 52-card deck, 4 cards pulled. Events: $A$: at least one club, $B$: at least one red card, Calculate $$C = A \cup B$$ probability. 2. Total ways: $$\binom{52}{4}.$$ 3. Calculate complements: $A^c$: no clubs chosen, so only 39 cards left. Ways: $$\binom{39}{4}.$$ $B^c$: no red cards chosen (26 black cards), ways: $$\binom{26}{4}.$$ 4. $A^c \cap B^c$ means no clubs and no red cards $ o$ only black non-club cards, which is 13 spades. Ways: $$\binom{13}{4}.$$ 5. Using inclusion-exclusion: $$P(C) = 1 - P(A^c \cap B^c) = 1 - \frac{\binom{13}{4}}{\binom{52}{4}}.$$ --- 1. **Problem 2.7:** From 35 cards each with one letter of the alphabet, pick 6 cards in order to form the word "энжжин" ("enzhin"). Find the probability of this exact word occurring. Also find the probability if their positions can be rearranged to form the word. 2. Since the word has length 6, probability of drawing those exact letters in order is: $$ \frac{\text{number of ways to pick those letters in exact order}}{\text{total ways to pick any 6 letters in order}}.$$ 3. Total ordered selections of 6 letters from 35: $$P(35,6) = \frac{35!}{29!}.$$ 4. The word "энжжин" has repeated letters "н" and "ж". Letters needed: count exactly to draw those letters with correct multiplicity; exactly one way to order. So favorable cases = 1. 5. Probability exact sequence: $$\frac{1}{P(35,6)} = \frac{29!}{35!}.$$ 6. If positions can be rearranged to form the word, the favorable cases = number of permutations of these letters: $$ \frac{6!}{2!2!} = 180.$$ 7. Probability: $$\frac{180}{P(35,6)} = \frac{180 \cdot 29!}{35!}.$$ --- 1. **Problem 2.8:** Along a line, $n$ boxes are placed side by side, $k$ balls are placed into these boxes randomly with equal probability for each ball to be in any box. Find the probability that each box has exactly one ball. 2. Assuming $k = n$, total ways to place balls into boxes independently: $n^k = n^n$. 3. Number of ways to place balls so that each box has exactly one ball is the number of permutations of $n$ balls into $n$ boxes: $$n!.$$ 4. Probability each box has exactly one ball: $$\frac{n!}{n^n}.$$