Question

a football squad contains 4 full - forwards, 6 full - backs and 3 centre - line players. find the probability of a full - forward and a centre - line player being chosen as captain and vice - captain, in any order.

Explanation:

Step 1: Find total number of players

The squad has 4 full - forwards, 6 full - backs and 3 centre - line players. So the total number of players \(n = 4+6 + 3=13\).

Step 2: Calculate the number of ways to choose a full - forward and a centre - line player (in any order)

The number of ways to choose a full - forward (4 choices) and a centre - line player (3 choices) in order is \(4\times3\). But since the order (captain and vice - captain) can be reversed (full - forward as vice - captain and centre - line as captain), we also have \(3\times4\) ways. But we can also think of it as the number of ways to choose 2 players where one is a full - forward and the other is a centre - line player. The number of ways to choose a full - forward and a centre - line player (in any order) is \(2\times(4\times3)\)? No, actually, the number of permutations of choosing 2 players where one is a full - forward (F) and one is a centre - line player (C) is \(P(4,1)\times P(3,1)+P(3,1)\times P(4,1)\)? Wait, no. The total number of permutations of choosing 2 players out of 13 is \(P(13,2)=13\times12\). The number of favorable permutations: if we first choose a full - forward (4 choices) and then a centre - line player (3 choices), that's \(4\times3\) ways. If we first choose a centre - line player (3 choices) and then a full - forward (4 choices), that's \(3\times4\) ways. So total favorable permutations \(=4\times3 + 3\times4=24\).

Alternatively, using probability:
Case 1: Captain is full - forward and Vice - captain is centre - line.
Probability that captain is full - forward: \(\frac{4}{13}\). Then, given that a full - forward is chosen as captain, the number of centre - line players is 3 and total remaining players is 12. So probability that vice - captain is centre - line: \(\frac{3}{12}=\frac{1}{4}\). So the probability for this case is \(\frac{4}{13}\times\frac{3}{12}=\frac{4\times3}{13\times12}=\frac{12}{156}=\frac{1}{13}\).

Case 2: Captain is centre - line and Vice - captain is full - forward.
Probability that captain is centre - line: \(\frac{3}{13}\). Then, given that a centre - line player is chosen as captain, the number of full - forwards is 4 and total remaining players is 12. So probability that vice - captain is full - forward: \(\frac{4}{12}=\frac{1}{3}\). So the probability for this case is \(\frac{3}{13}\times\frac{4}{12}=\frac{12}{156}=\frac{1}{13}\).

Total probability \(=\frac{1}{13}+\frac{1}{13}=\frac{2}{13}\)? Wait, no. Wait, \(\frac{4}{13}\times\frac{3}{12}=\frac{12}{156}=\frac{1}{13}\), \(\frac{3}{13}\times\frac{4}{12}=\frac{12}{156}=\frac{1}{13}\). So total probability \(=\frac{1 + 1}{13}=\frac{2}{13}\)? Wait, no, \(\frac{4\times3+3\times4}{13\times12}=\frac{24}{156}=\frac{2}{13}\). Yes, that's correct.

Wait, let's recalculate:
Total number of players \(N = 4 + 6+3=13\).
Number of ways to choose 2 players (captain and vice - captain) where one is a full - forward (F) and one is a centre - line (C) player:

• First, choose F then C: \(4\times3 = 12\) ways.
• First, choose C then F: \(3\times4=12\) ways.

Total favorable ways \(=12 + 12 = 24\) ways.
Total number of ways to choose 2 players out of 13 (captain and vice - captain) is \(13\times12=156\) ways.
So probability \(=\frac{24}{156}=\frac{2}{13}\).

Answer:

\(\frac{2}{13}\)