There are 12 people in total, and they are divided into groups of 5, 4, and 3. We need to find the probability that Jai and Kaz are in the same group.

Method 1: Probabilities

Calculate the probability for each group:

• In the group of 5: \(\frac{5}{12} \times \frac{4}{11} = \frac{20}{132} = \frac{5}{33}\)
• In the group of 4: \(\frac{4}{12} \times \frac{3}{11} = \frac{12}{132} = \frac{1}{11}\)
• In the group of 3: \(\frac{3}{12} \times \frac{2}{11} = \frac{6}{132} = \frac{1}{22}\)

Add the probabilities for the three scenarios:

\(\frac{5}{33} + \frac{1}{11} + \frac{1}{22} = \frac{19}{66}\)

Method 2: Arrangements

Calculate the number of ways Jai and Kaz can be in each group:

• In the group of 5: \(\binom{10}{3} \times \binom{7}{4} = 4200\)
• In the group of 4: \(\binom{10}{2} \times \binom{8}{5} = 2520\)
• In the group of 3: \(\binom{10}{1} \times \binom{9}{5} = 1260\)

Total number of ways to arrange the groups: \(\binom{12}{5} \times \binom{7}{4} = 27720\)

Probability: \(\frac{4200 + 2520 + 1260}{27720} = \frac{7980}{27720} = \frac{19}{66}\)