Question

In a school, the probability of holding a debate competition is $\frac{1}{3}$ and that of a quiz competition is $\frac{2}{3}$. In the two participating teams, A has 4 girls and 6 boys and B has 7 girls and 3 boys. If a debate competition is held, the students are selected from team A and for the quiz competition they are selected from team B. If only two students are to be chosen from the teams, then find the probability that one will be a girl and the other a boy.

Hint:

Keep the denominators common ($\frac{135}{}$) throughout your intermediate calculations instead of simplifying $P(E|D)$ and $P(E|Q)$ too early. This will make adding fractions at the end much simpler.

Answer 1

Concept: This problem can be resolved through the application of the Total Probability Theorem. There are two mutually exclusive events representing the choice of competition, and each competition dictates selection from a specific composition of students.
• Total Probability Formula: $P(E) = P(C_1) \cdot P(E|C_1) + P(C_2) \cdot P(E|C_2)$
• Combinatorics Component: The number of ways to pick 1 girl and 1 boy from a group of $G$ girls and $B$ boys when selecting 2 students is given by $\frac{\binom{G}{1} \times \binom{B}{1}}{\binom{G+B}{2}}$.

Step 1: Calculate the selection probability assuming the debate competition is held (Team A).
Let $D$ be the event that a debate competition is held, and $Q$ be the event that a quiz competition is held. From the problem description: \[ P(D) = \frac{1}{3}, \quad P(Q) = \frac{2}{3} \] If a debate competition occurs, two students are picked from Team A (4 girls, 6 boys; Total students = 10). Let $E$ be the event that 1 girl and 1 boy are selected. The conditional probability $P(E|D)$ is calculated as: \[ P(E|D) = \frac{\binom{4}{1} \times \binom{6}{1}}{\binom{10}{2}} = \frac{4 \times 6}{\frac{10 \times 9}{2}} = \frac{24}{45} \]

Step 2: Calculate the selection probability assuming the quiz competition is held (Team B) and apply the total probability rule.
If a quiz competition occurs, two students are picked from Team B (7 girls, 3 boys; Total students = 10). The conditional probability $P(E|Q)$ is calculated as: \[ P(E|Q) = \frac{\binom{7}{1} \times \binom{3}{1}}{\binom{10}{2}} = \frac{7 \times 3}{\frac{10 \times 9}{2}} = \frac{21}{45} \] Now, apply the Law of Total Probability to find the total probability $P(E)$: \[ P(E) = P(D) \cdot P(E|D) + P(Q) \cdot P(E|Q) \] Substitute the calculated fractions into the formula: \[ P(E) = \left( \frac{1}{3} \times \frac{24}{45} \right) + \left( \frac{2}{3} \times \frac{21}{45} \right) \] \[ P(E) = \frac{24}{135} + \frac{42}{135} = \frac{66}{135} \] Simplify the final fraction by dividing the numerator and denominator by their greatest common divisor, which is 9: \[ P(E) = \frac{66 \div 3}{135 \div 3} = \frac{22}{45} \] Let's re-verify the arithmetic: $24/135 + 42/135 = 66/135$. Dividing by 3 gives $22/45$.