Question

Suppose that 35 people are divided in a random manner into two teams in such a way that one team contains 10 people and the other team contains 25 people. What is the probability that two particular people A and B will be on the same team?

Answer 1

Answer:

0.5798 or 57.98%

Step-by-step explanation:

The total number of ways to form the two teams is the combination of choosing 10 people out of 35 (₃₅C₁₀). The number of possibilities that A and B are both on the 10-people team is given by the combination of choosing 8 people (since two are fixed) out of 33 (₃₃C₈).The number of possibilities that A and B are both on the 25-people team is given by the combination of choosing 10 people out of 33 (₃₃C₈).

Therefore, the probability that two particular people A and B will be on the same team is:

$P = \frac{_{33}C_8+_{33}C_{10}}{_{35}C_{10}} \\\\P=\frac{\frac{33!}{(33-8)!8!}+\frac{33!}{(33-10)!10!} }{\frac{35!}{(35-10)!10!}} \\\\P=\frac{69}{119} = 0.5798$

The probability is 0.5798 or 57.98%.