A committee of 4 people is to be chosen from 6 men and 5 women. What is the probability that the committee includes at least 2 women?
["Title: Probability of Selecting a Committee with At Least 2 Women from 6 Men and 5 Women", "When forming a decision-making or representative group such as a committee, understanding the probability of specific gender composition plays a vital role—especially in settings promoting diversity and fairness. This article explores a classic combinatorics problem: selecting a 4-person committee from 6 men and 5 women, and calculating the probability that the committee includes at least 2 women.", "---", "### Problem Overview", "We are given:\n- 6 men\n- 5 women\n- A committee of exactly 4 people must be chosen randomly from the total of 11 individuals (6 + 5).", "We want to find:\nWhat is the probability that the selected committee includes at least 2 women?", "---", "### Step 1: Total Number of Ways to Choose a Committee of 4", "From a group of 11 people, the total number of ways to form a committee of 4 is given by the combination formula:", "[\n\binom{11}{4} = \frac{11!}{4!(11-4)!} = \frac{11 \ imes 10 \ imes 9 \ imes 8}{4 \ imes 3 \ imes 2 \ imes 1} = 330\n]", "So, there are 330 possible ways to form any 4-person committee.", "---", "### Step 2: Define "At Least 2 Women"", ""The committee includes at least 2 women" means possible gender distributions are:\n- Exactly 2 women and 2 men\n- Exactly 3 women and 1 man\n- Exactly 4 women and 0 men", "We will calculate the number of favorable outcomes for each of these cases and sum them up.", "---", "### Step 3: Count Favorable Outcomes", "Case 1: 2 women and 2 men\nChoose 2 women from 5:\n[\n\binom{5}{2} = \frac{5 \ imes 4}{2 \ imes 1} = 10\n]\nChoose 2 men from 6:\n[\n\binom{6}{2} = \frac{6 \ imes 5}{2 \ imes 1} = 15\n]\nTotal for this case:\n[\n10 \ imes 15 = 150\n]", "Case 2: 3 women and 1 man\nChoose 3 women from 5:\n[\n\binom{5}{3} = \binom{5}{2} = 10 \quad \ ext{(since } \binom{n}{k} = \binom{n}{n-k}\ ext{)}\n]\nChoose 1 man from 6:\n[\n\binom{6}{1} = 6\n]\nTotal for this case:\n[\n10 \ imes 6 = 60\n]", "Case 3: 4 women and 0 men\nChoose 4 women from 5:\n[\n\binom{5}{4} = 5\n]\nNo men needed, so only 5 ways.", "---", "### Step 4: Total Favorable Outcomes", "Add all favorable cases:\n[\n150 + 60 + 5 = 215\n]", "---", "### Step 5: Compute the Probability", "[\n\ ext{Probability} = \frac{\ ext{Favorable outcomes}}{\ ext{Total outcomes}} = \frac{215}{330}\n]", "Simplify the fraction:\nDivide numerator and denominator by 5:\n[\n\frac{215 \div 5}{330 \div 5} = \frac{43}{66}\n]", "---", "### Final Answer", "The probability that the committee includes at least 2 women is:", "[\n\boxed{\frac{43}{66}}\n]", "---", "### Why This Matters", "Understanding such probabilities helps organizations ensure balanced representation in committees, boards, or panels. By calculating exact chances based on inclusive sampling rules, decision-makers can foster fairness and meet diversity goals with concrete statistical backing.", "If you're involved in planning balanced teams, committees, or selection processes, analyzing these probabilities ensures informed, equitable outcomes."]
