A box contains 25 identical balls, 10 of them are blad, Tof them are red and the restare white: If thee balls are drawn at random from a box pre after the sther wishout eplacement ifind the probabilythat a. one is blact one is red and one is shite b. exactly two are black c. none of them is red.

We are given a box with 25 balls. According to the problem, there are:   • 10 black balls,   • 2 red balls, and   • the rest are white. Since the total is 25, the number of white balls is   25 – (10 + 2) = 13 white balls. We draw three balls at random without replacement. The total number of possible draws is   Total outcomes = C(25, 3). Recall that   C(n, k) = n!/(k!(n – k)!). Thus,   C(25, 3) = 25×24×23 ÷ 6 = 2300. Below are the solutions for each part: ───────────────────────────── Part (a): One black, one red, and one white Number of ways to choose:   • 1 black out of 10:  C(10, 1)   • 1 red out of 2:   C(2, 1)   • 1 white out of 13:  C(13, 1) Since these selections are independent, the number of favorable outcomes is   Favorable outcomes = C(10, 1) × C(2, 1) × C(13, 1) = 10 × 2 × 13 = 260. Thus, the required probability is   P(a) = 260 / 2300. You can simplify it by dividing numerator and denominator by 10:   P(a) = 26/230 = 13/115. ───────────────────────────── Part (b): Exactly two are black Here we want exactly 2 black balls and 1 ball that is not black (that is, either red or white). Step 1. Choose 2 blacks out of 10:   C(10, 2). Step 2. Choose 1 non-black from the remaining balls. The non-black balls total   2 (red) + 13 (white) = 15. So, choose 1 out of 15:  C(15, 1). The number of favorable outcomes is   Favorable outcomes = C(10, 2) × C(15, 1). Now, C(10, 2) = 10×9/2 = 45, so   Favorable outcomes = 45 × 15 = 675. Thus, the probability is   P(b) = 675 / 2300. You can simplify this fraction if desired. Dividing numerator and denominator by 5 gives:   675 ÷ 5 = 135  and 2300 ÷ 5 = 460, so   P(b) = 135/460. Dividing numerator and denominator by 5 again:   135 ÷ 5 = 27  and 460 ÷ 5 = 92, thus   P(b) = 27/92. ───────────────────────────── Part (c): None of them is red For no red balls, all three balls must be chosen from the non-red balls. The number of non-red balls is   10 (black) + 13 (white) = 23. So, the number of favorable outcomes is   Favorable outcomes = C(23, 3). Thus, the probability is   P(c) = C(23, 3) / C(25, 3). To compute C(23, 3):   C(23, 3) = 23×22×21 ÷ 6 = 1771. So,   P(c) = 1771 / 2300. ───────────────────────────── Summary of Answers: a. P(one black, one red, one white) = [10×2×13] / C(25, 3) = 260/2300 = 13/115 b. P(exactly two black) = [C(10, 2) × C(15, 1)] / C(25, 3) = 675/2300 = 27/92 c. P(no red) = C(23, 3) / C(25, 3) = 1771/2300 These are the final answers for the problem.