Understanding the Problem
In probability theory, when we choose items from a group without replacing them, we are dealing with dependent events. In this scenario, we have a class of 100 students: 60 boys and 40 girls. We need to find the probability that choosing two students results in one boy and one girl.
The Calculation Approach
There are two main ways to approach this: using combinations or calculating individual branch probabilities.
Method 1: Using Combinations
We want to choose 1 boy from 60 and 1 girl from 40. The total number of ways to choose any 2 students from 100 is given by the combination formula $_nC_r = \frac{n!}{r!(n-r)!}$.
- Total outcomes: $\binom{100}{2} = \frac{100 \times 99}{2} = 4950$
- Favorable outcomes (1 boy AND 1 girl): $\binom{60}{1} \times \binom{40}{1} = 60 \times 40 = 2400$
- Probability: $P = \frac{2400}{4950} = \frac{240}{495} = \frac{48}{99} = \frac{16}{33} \approx 0.4848$
Method 2: Sequential Probabilities
We can pick a boy first, then a girl, OR a girl first, then a boy.
- Scenario A (Boy then Girl): $P(B_1 \cap G_2) = \frac{60}{100} \times \frac{40}{99} = \frac{2400}{9900} = \frac{24}{99}$
- Scenario B (Girl then Boy): $P(G_1 \cap B_2) = \frac{40}{100} \times \frac{60}{99} = \frac{2400}{9900} = \frac{24}{99}$
Adding these disjoint scenarios together: $\frac{24}{99} + \frac{24}{99} = \frac{48}{99} = \frac{16}{33}$.
Intuition
Because we are picking two people, the first choice changes the probability for the second choice (the total number of students drops from 100 to 99). Always ensure you account for both possible orders (BG and GB) when the prompt doesn't specify a strict order.