answer:To solve this problem, we first need to understand the concept of combinations, denoted as C_n^r, which represents the number of ways to choose r items out of n without regard to the order. Given that there are 3 good products and 2 defective ones among the 5 products, and we are selecting 2 products, we can break down the probability of selecting exactly 1 good product into steps: 1. Calculate the number of ways to select 1 good product out of 3. This can be done using the combination formula C_n^r = frac{n!}{r!(n-r)!}, where n is the total number of items, and r is the number of items to choose. For selecting 1 good product, it is C_3^1 = frac{3!}{1!(3-1)!} = 3. 2. Calculate the number of ways to select 1 defective product out of 2. Using the combination formula again, it is C_2^1 = frac{2!}{1!(2-1)!} = 2. 3. Calculate the total number of ways to select 2 products out of 5, which is C_5^2 = frac{5!}{2!(5-2)!} = 10. 4. The probability of selecting exactly 1 good product is the product of the number of ways to select 1 good and 1 defective product divided by the total number of ways to select 2 products from 5. This gives us P = frac{C_3^1 cdot C_2^1}{C_5^2} = frac{3 cdot 2}{10} = frac{6}{10} = frac{3}{5}. Therefore, the probability of selecting exactly 1 good product when 2 products are randomly selected from 5 products, with 3 being good and 2 being defective, is boxed{frac{3}{5}}.

answer:The problem can be analyzed by considering the various possible clockwise orderings of the points around the circle. For chords overline{PQ} and overline{RS} to intersect, the points must not all appear sequentially when placed in clockwise order. Specifically, two points must intersperse between the two points of the other chord. We consider the six possible orderings of four points: 1. PRQS 2. PSRQ 3. PQRS 4. PSQR 5. PQSR 6. PRSQ Out of these, the orderings where overline{PQ} and overline{RS} intersect are: 1. PRQS 2. PSRQ In these cases, one point of each chord is between the points of the other chord, satisfying the condition for intersection. Since all orderings are equally likely, the probability that overline{PQ} intersects overline{RS} is frac{2}{6} = boxed{frac{1}{3}}.

answer:Part (a): Prove that at some point during the game, the time difference between the clocks was at least 1 minute 51 seconds. 1. Let the notation Delta(t) represent the absolute difference in the times displayed by the two clocks at any given moment t during the game. 2. Suppose for contradiction that Delta(t) < 1 minute 51 seconds left(frac{111}{60} text{ minutes}right) at all times during the game. 3. Since after both players have made 40 moves, the total time spent by each player is 2 hours 30 minutes (150 minutes) and both clocks show the same time at the end: [ Delta(0) = 0 quad text{and} quad Delta(text{end}) = 0 quad (text{at } 150 text{ minutes}) ] 4. Note that making one move switches the active player's clock to the opponent’s clock. Thus, if we denote Delta(t) as the time difference right after player 1's move and right before player 2 starts their move, we have: [ 0 leq Delta(t) < frac{111}{60} text{ minutes} ] 5. The total duration over which we can accumulate these differences Delta per move over 40 moves of each player (a total of 80 segments): [ 80 times frac{111}{60} = frac{8880}{60} = 148 text{ minutes} ] 6. This is a contradiction since the total game time is fixed at 150 minutes, but we supposedly never saw a difference greater than frac{37}{20} approx 1 minute 51 seconds. Thus, there must be at least one moment in the game where the difference between the two clocks was at least 1 minute and 51 seconds. boxed{} Part (b): Can it be asserted that at some moment, the difference between the clocks was exactly 2 minutes? 1. Let's demonstrate a setup where the absolute difference in times shown by clocks never exceeds 2 minutes: 2. Consider an example where: - Player 1 uses time tau = frac{300}{158} minutes for the first move. - Each subsequent move by either player requires 2tau minutes, except the last move by Player 2 who spends some remaining time to total exactly 150 minutes. 3. Thus: - For 40 total moves including both, for each k-th move, the difference Delta(t) = |t_i - (t_{i+k})| where t_i, t_{i+k} are the times taken by players: [ 79 times 2tau + tau = 150 text{ minutes} ] - Each k tau < 2 minutes. Therefore, we can construct a situation where the time difference recorded at any instance is less than 2 minutes. Hence, we cannot generally assert a definite moment where the difference equals precisely 2 minutes. boxed{text{No}}

answer:To solve for the coefficient of x^3 in the expression (x+1)^{50}: 1. Identify the term in the expanded formula that contains x^3. This term can be represented as binom{50}{3} cdot x^3 cdot 1^{47}. 2. Calculate binom{50}{3}: [ binom{50}{3} = frac{50 times 49 times 48}{3 times 2 times 1} = frac{117600}{6} = 19600. ] 3. Therefore, the coefficient of x^3 term is boxed{19600}.