1. **Problem Statement:** Three coins are tossed simultaneously. Find the probability of: (i) Exactly two heads (ii) At least two heads (iii) At most two heads 2. **Total possible outcomes:** Each coin has 2 outcomes (Head or Tail), so total outcomes = $2^3 = 8$. 3. **Calculate probabilities:** (i) Exactly two heads means 2 heads and 1 tail. Possible outcomes: HHT, HTH, THH (3 outcomes). Probability = $\frac{3}{8}$. (ii) At least two heads means 2 or 3 heads. Number of outcomes with 3 heads: HHH (1 outcome). Total outcomes with at least two heads = 3 (exactly two) + 1 (three heads) = 4. Probability = $\frac{4}{8} = \frac{1}{2}$. (iii) At most two heads means 0, 1, or 2 heads. Number of outcomes with 0 heads: TTT (1 outcome). Number of outcomes with 1 head: HTT, THT, TTH (3 outcomes). Number of outcomes with 2 heads: 3 (from above). Total = 1 + 3 + 3 = 7. Probability = $\frac{7}{8}$. --- 4. **Problem Statement:** Find the perimeter of the shaded region where ADC, AEB, and BFC are semicircles on diameters AC, AB, and BC respectively. 5. **Given:** - ADC is the largest semicircle on diameter AC. - AEB and BFC are smaller semicircles on diameters AB and BC. - The shaded region is inside ADC but outside AEB and BFC. 6. **Approach:** The perimeter of the shaded region consists of the arcs of the two smaller semicircles AEB and BFC (outside the shaded region) and the arc of the largest semicircle ADC (inside the shaded region). 7. **Formula for semicircle perimeter (arc length):** $$\text{Arc length} = \pi \times r$$ where $r$ is the radius of the semicircle. 8. **Let lengths:** - $AB = a$ - $BC = b$ - $AC = a + b$ 9. **Calculate perimeter:** Perimeter of shaded region = Arc of ADC - (Arcs of AEB + BFC) $$= \pi \times \frac{a+b}{2} - \pi \times \frac{a}{2} - \pi \times \frac{b}{2} = 0$$ This means the perimeter of the shaded region is the sum of the arcs of the two smaller semicircles. 10. **Median of given data:** Data intervals and frequencies: 75-84: 8 85-94: 11 95-104: 26 105-114: 31 115-124: 18 125-134: 4 135-144: 2 Total frequency $N = 8+11+26+31+18+4+2 = 100$. Median class is where cumulative frequency crosses $\frac{N}{2} = 50$. Cumulative frequencies: - 75-84: 8 - 85-94: 19 - 95-104: 45 - 105-114: 76 (median class) Median class = 105-114 Median formula: $$\text{Median} = L + \left(\frac{\frac{N}{2} - F}{f}\right) \times h$$ where $L = 105$ (lower boundary of median class), $F = 45$ (cumulative frequency before median class), $f = 31$ (frequency of median class), $h = 10$ (class width). Calculate: $$\text{Median} = 105 + \left(\frac{50 - 45}{31}\right) \times 10 = 105 + \frac{5}{31} \times 10 = 105 + 1.61 = 106.61$$ **Final answers:** (i) Probability exactly two heads = $\frac{3}{8}$ (ii) Probability at least two heads = $\frac{1}{2}$ (iii) Probability at most two heads = $\frac{7}{8}$ Perimeter of shaded region = sum of arcs of smaller semicircles = $\pi \times \frac{a}{2} + \pi \times \frac{b}{2} = \frac{\pi}{2}(a+b)$ Median of data = 106.61