your employer is offering free tickets for two basketball games. They play 60 home games and six of them are against the snakes. The tickets are randomly given out. What is the probability that both games are against the snakes?
The probability of randomly choosing one game against the snakes out of the 60 home games is 6/60 = 1/10.
To find the probability of both games being against the snakes, we multiply the probability of the first game being against the snakes by the probability of the second game also being against the snakes:
(1/10) * (1/10) = 1/100
Therefore, the probability that both games are against the snakes is 1/100.