Probability of Selecting 2 Men from a Group
This section details the calculation for the probability of selecting exactly 2 men when 3 people are chosen randomly from a mixed group.
Group Composition Overview
The total group comprises:
- Number of Men: 4
- Number of Women: 3
- Number of Boys: 2
- Total Persons: $4 + 3 + 2 = 9$
Total Selections Calculation
The total number of ways to select 3 persons from 9 is found using combinations ($C(n, k) = \frac{n!}{k!(n-k)!}$):
Total ways = $C(9, 3) = \frac{9!}{3!(9-3)!} = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} = 84$
Favorable Selections Calculation
Favorable outcomes require selecting exactly 2 men and 1 other person (non-man).
- Ways to choose 2 men out of 4:
$C(4, 2) = \frac{4!}{2!2!} = \frac{4 \times 3}{2 \times 1} = 6$
- Ways to choose 1 person from the remaining 5 (3 women + 2 boys):
$C(5, 1) = \frac{5!}{1!4!} = 5$
- Total favorable ways = (Ways to choose men) $\times$ (Ways to choose other)
$6 \times 5 = 30$
Probability Calculation Step
Probability is calculated as the ratio of favorable outcomes to total possible outcomes:
Probability = $\frac{\text{Favorable Ways}}{\text{Total Ways}} = \frac{30}{84}$
Simplifying the fraction gives:
$\frac{30 \div 6}{84 \div 6} = \frac{5}{14}$
Final Probability Result
The probability of selecting exactly 2 men is $\frac{5}{14}$.
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