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.
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.
Show Hint
• Always ensure the sum of probabilities of the mutually exclusive events (Debate and Quiz) is 1.
• For selection of two items of different types, use the product of individual combinations divided by the total combination.
• For selection of two items of different types, use the product of individual combinations divided by the total combination.
Updated On: Sep 10, 2026
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Solution and Explanation
Concept:
• Law of Total Probability: \( P(E) = P(A_1)P(E|A_1) + P(A_2)P(E|A_2) \).
• Combination formula for selection: \( ^nC_r = \frac{n!}{r!(n-r)!} \).
Step 1: Define the events and their probabilities
Let \( D \) be the event that a debate competition is held, and \( Q \) be the event that a quiz competition is held.
\( P(D) = \frac{1}{3} \)
\( P(Q) = \frac{2}{3} \)
Let \( E \) be the event that one girl and one boy are selected.
Step 2: Calculate conditional probabilities for each competition
For Debate (Team A: 4G, 6B, Total 10):
The number of ways to select 1 girl and 1 boy out of 2 students is:
\( P(E|D) = \frac{^4C_1 \times ^6C_1}{^{10}C_2} = \frac{4 \times 6}{45} = \frac{24}{45} \)
For Quiz (Team B: 7G, 3B, Total 10):
The number of ways to select 1 girl and 1 boy out of 2 students is:
\( P(E|Q) = \frac{^7C_1 \times ^3C_1}{^{10}C_2} = \frac{7 \times 3}{45} = \frac{21}{45} \)
Step 3: Apply the Law of Total Probability
\( P(E) = P(D)P(E|D) + P(Q)P(E|Q) \)
\( P(E) = \left( \frac{1}{3} \times \frac{24}{45} \right) + \left( \frac{2}{3} \times \frac{21}{45} \right) \)
\( P(E) = \frac{24}{135} + \frac{42}{135} = \frac{66}{135} \)
Simplifying the fraction by dividing by 3:
\( P(E) = \frac{22}{45} \)
• Law of Total Probability: \( P(E) = P(A_1)P(E|A_1) + P(A_2)P(E|A_2) \).
• Combination formula for selection: \( ^nC_r = \frac{n!}{r!(n-r)!} \).
Step 1: Define the events and their probabilities
Let \( D \) be the event that a debate competition is held, and \( Q \) be the event that a quiz competition is held.
\( P(D) = \frac{1}{3} \)
\( P(Q) = \frac{2}{3} \)
Let \( E \) be the event that one girl and one boy are selected.
Step 2: Calculate conditional probabilities for each competition
For Debate (Team A: 4G, 6B, Total 10):
The number of ways to select 1 girl and 1 boy out of 2 students is:
\( P(E|D) = \frac{^4C_1 \times ^6C_1}{^{10}C_2} = \frac{4 \times 6}{45} = \frac{24}{45} \)
For Quiz (Team B: 7G, 3B, Total 10):
The number of ways to select 1 girl and 1 boy out of 2 students is:
\( P(E|Q) = \frac{^7C_1 \times ^3C_1}{^{10}C_2} = \frac{7 \times 3}{45} = \frac{21}{45} \)
Step 3: Apply the Law of Total Probability
\( P(E) = P(D)P(E|D) + P(Q)P(E|Q) \)
\( P(E) = \left( \frac{1}{3} \times \frac{24}{45} \right) + \left( \frac{2}{3} \times \frac{21}{45} \right) \)
\( P(E) = \frac{24}{135} + \frac{42}{135} = \frac{66}{135} \)
Simplifying the fraction by dividing by 3:
\( P(E) = \frac{22}{45} \)
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