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 full - back being chosen as captain and vice - captain, in any order.

Explanation:

Step1: Calculate total number of players

Total players = 4 (full - forwards) + 6 (full - backs) + 3 (centre - line players) = 13.

Step2: Calculate probability of full - forward then full - back

Probability of choosing a full - forward first: $\frac{4}{13}$. Then, since one player is already chosen (non - replacement), total players left is 12, and full - backs left is 6. Probability of choosing a full - back next: $\frac{6}{12}=\frac{1}{2}$. So the probability of full - forward then full - back is $\frac{4}{13}\times\frac{6}{12}=\frac{4}{13}\times\frac{1}{2}=\frac{2}{13}$.

Step3: Calculate probability of full - back then full - forward

Probability of choosing a full - back first: $\frac{6}{13}$. Then, total players left is 12, and full - forwards left is 4. Probability of choosing a full - forward next: $\frac{4}{12}=\frac{1}{3}$. So the probability of full - back then full - forward is $\frac{6}{13}\times\frac{4}{12}=\frac{6}{13}\times\frac{1}{3}=\frac{2}{13}$.

Step4: Calculate total probability

Since the two events (full - forward then full - back and full - back then full - forward) are mutually exclusive, we add their probabilities. Total probability = $\frac{2}{13}+\frac{2}{13}=\frac{4}{13}$. Wait, no, wait. Wait, let's re - calculate. Wait, when we choose full - forward first: number of full - forwards is 4, total is 13. Then full - back: 6 out of 12. So $\frac{4}{13}\times\frac{6}{12}=\frac{4\times6}{13\times12}=\frac{24}{156}=\frac{2}{13}$. When we choose full - back first: 6 out of 13, then full - forward: 4 out of 12. So $\frac{6}{13}\times\frac{4}{12}=\frac{24}{156}=\frac{2}{13}$. Then total probability is $\frac{2}{13}+\frac{2}{13}=\frac{4}{13}$? Wait, no, wait another way. The number of ways to choose a full - forward and a full - back in any order: (number of full - forwards×number of full - backs)×2 (for order) divided by (total number of ways to choose 2 players from 13). The number of ways to choose 2 players from 13 is $C(13,2)=\frac{13!}{2!(13 - 2)!}=\frac{13\times12}{2\times1}=78$. The number of ways to choose a full - forward and a full - back: 4×6×2 = 48 (4 full - forwards, 6 full - backs, and 2 orders). So probability is $\frac{48}{78}=\frac{8}{13}$? Wait, no, I made a mistake earlier. Wait, 4 full - forwards, 6 full - backs. The number of pairs with one full - forward and one full - back is 4×6×2? No, no. The number of ways to choose one full - forward and one full - back (unordered) is 4×6 = 24. The number of ways to choose any two players is $C(13,2)=\frac{13\times12}{2}=78$. So probability is $\frac{24\times2}{78}$? No, no. Wait, in ordered pairs: the number of ordered pairs with full - forward first and full - back second is 4×6 = 24. The number of ordered pairs with full - back first and full - forward second is 6×4 = 24. Total ordered pairs of two players: 13×12 = 156. So total probability is $\frac{24 + 24}{156}=\frac{48}{156}=\frac{4}{13}$. Ah, so that's correct. So the probability is $\frac{4}{13}$.

Answer:

$\frac{4}{13}$ (corresponding to the option with $\frac{4}{13}$)