Question

The probability of an event - picking teams. About (a) 10 kids are randomly grouped into an A team with five kids and a B team with five kids. Each grouping is equally likely. There are three kids in the group, Alex and his two best friends Jose and Carl. What is the probability that Alex ends up on the same team with at least one of his two best friends?

Answer 1

Answer: The required probability is $\dfrac{4}{9}$

Step-by-step explanation:

Since we have given that

Number of kids = 10

Number of kids in Team A = 5

Number of kids in Team B = 5

There are three kids in the group, Alex and his two best friends Jose and Carl.

So, number of favourable outcome is given by

$2(\dfrac{8!}{3!\times 5!})$

Total number of outcomes is given by

$\dfrac{10!}{5!\times 5!}$

So, the probability that Alex ends up on the same team with at least one of his two best friends is given by

$\dfrac{2(\dfrac{8!}{5!\times 3!)}}{\dfrac{10!}{5!\times 5!}}\\\\=2\times \dfrac{8!}{3!}\times \dfrac{5!}{10!}\\\\=\dfrac{4}{9}$

Hence, the required probability is $\dfrac{4}{9}$