There are a total of 54 games, so there are 54 options for the first game. Since there are 7 games against the Snakes, there are 7 options for the first game to be against the Snakes.
After the first game, there are only 53 games left, so there are 53 options for the second game. Likewise, there are 6 games against the Snakes left, so there are 6 options for the second game to be against the Snakes.
By the multiplication principle (assuming independence), the probability that both games are against the Snakes is $\dfrac{7}{54} \times \dfrac{6}{53} = \dfrac{42}{2862} = \boxed{\dfrac{7}{477}}$.
Your employer is offering free tickets for two baseball games. They play
54
54 home games and
7
7 of them are against the Snakes. The tickets are randomly given out. What is the probability that both are games are against the Snakes? Write your answer as a fraction. You do not need to simplify.
1 answer