Question

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

Explanation:

Step1: Calculate total number of players

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

Step2: Case 1: Captain is full - back, Vice - captain is centre - line player

Probability (Captain: full - back, Vice - captain: centre - line) = \(\frac{6}{13}\times\frac{3}{12}=\frac{6\times3}{13\times12}=\frac{18}{156}=\frac{3}{26}\)

Step3: Case 2: Captain is centre - line player, Vice - captain is full - back

Probability (Captain: centre - line, Vice - captain: full - back) = \(\frac{3}{13}\times\frac{6}{12}=\frac{3\times6}{13\times12}=\frac{18}{156}=\frac{3}{26}\)

Step4: Total probability

Total probability = \(\frac{3}{26}+\frac{3}{26}=\frac{6}{26}=\frac{3}{13}\)

Answer:

\(\frac{3}{13}\) (the option with \(\boldsymbol{\frac{3}{13}}\))