answer:From the problem statement, the probability of drawing a red ball at each draw is frac{3}{8}. For X=3, the first two draws must consist of one red ball and one white ball in any order, and the third draw must be a red ball to satisfy the condition of stopping at two red balls. The number of ways to arrange one red and one white ball in two draws is C_2^1, which is 2. The probabilities for the draws are: - First draw: White ball with probability frac{5}{8}, red ball with probability frac{3}{8}. - Second draw: If the first was white, then we need a red, again frac{3}{8}; if the first was red, then we need a white, which is frac{5}{8}. - Third draw must be a red ball with probability frac{3}{8}. Combining these, we get the probability P(X=3) as the product of the three draws with the combinations: P(X=3) = C_2^1 cdot left(frac{5}{8} cdot frac{3}{8}right) cdot frac{3}{8} + C_2^1 cdot left(frac{3}{8} cdot frac{5}{8}right) cdot frac{3}{8} Since both terms are the same, we can simplify the calculation: P(X=3) = 2 cdot left(frac{5}{8} cdot frac{3}{8}right) cdot frac{3}{8} = 2 cdot frac{15}{64} cdot frac{3}{8} = frac{30}{64} cdot frac{3}{8} Simplifying further: P(X=3) = frac{30}{64} cdot frac{3}{8} = frac{90}{512} = frac{45}{256} Therefore, the final answer is: boxed{P(X=3) = frac{45}{256}}

answer:To find out how many pencils Mike placed in the drawer, you subtract the original number of pencils from the total number of pencils after Mike added his. So, you calculate 71 (total pencils) - 41 (original pencils) = 30 pencils. Mike placed boxed{30} pencils in the drawer.

answer:Solution: (Ⅰ) Let A(x_0, 0), B(0, y_0), P(x, y), From overrightarrow{BP} = 2overrightarrow{PA}, we get (x, y-y_0) = 2(x_0 - x, -y), which implies begin{cases} x = 2(x_0 - x) y - y_0 = -2y end{cases} Rightarrow begin{cases} x_0 = frac{3}{2}x y_0 = 3y end{cases}, Since x_0^2 + y_0^2 = 9, we have left(frac{3}{2}xright)^2 + (3y)^2 = 9, Simplifying gives: frac{x^2}{4} + y^2 = 1, which is the equation of the trajectory of point P. (Ⅱ) When the line passing through (1, 0) is y = 0, overrightarrow{OM} cdot overrightarrow{ON} = (2, 0) cdot (-2, 0) = -4, When the line passing through (1, 0) is not y = 0, it can be set as x = ty + 1, let A(x_1, y_1), B(x_2, y_2), By combining frac{x^2}{4} + y^2 = 1 and x = ty + 1, we get: (t^2 + 4)y^2 + 2ty - 3 = 0, By Vieta's formulas, we have y_1 + y_2 = -frac{2t}{t^2 + 4}, y_1y_2 = -frac{3}{t^2 + 4}, overrightarrow{OM} cdot overrightarrow{ON} = x_1x_2 + y_1y_2 = (ty_1 + 1)(ty_2 + 1) + y_1y_2 = (t^2 + 1)y_1y_2 + t(y_1 + y_2) + 1 = frac{-4t^2 + 1}{t^2 + 4} = frac{-4(t^2 + 4) + 17}{t^2 + 4} = -4 + frac{17}{t^2 + 4} Since Delta = 4t^2 + 12(t^2 + 4) = 16t^2 + 48 > 0 always holds, we have t in mathbb{R}, and for the above expression, when t = 0, left(overrightarrow{OM} cdot overrightarrow{ON}right)_{text{max}} = frac{1}{4} In summary, the maximum value of overrightarrow{OM} cdot overrightarrow{ON} is boxed{frac{1}{4}}.

answer:The correct answer is C. The detailed solution is omitted. Therefore, the equation of the perpendicular bisector of AB is boxed{text{C: } 3x - y - 9 = 0}.