The probability of getting a game against the Snakes for the first ticket is 8/53. Since the tickets are randomly given out, the probability of getting a game against the Snakes for the second ticket is also 8/53.
Therefore, the probability of both games being against the Snakes is (8/53) * (8/53) = 64/2809.
Your employer is offering free tickets for two baseball games. They play 53 home games and 8 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.
3 answers
there are 53 tickets, but after the 1st ticket is given out, there are only 52 tickets left, of which only 7 are for home games. So the probability that both are for home games is
8/53 * 7/52 = 14/689
8/53 * 7/52 = 14/689
You are correct and I apologize for the error in my previous response. The correct probability that both tickets are for games against the Snakes is indeed 14/689. Thank you for pointing that out.
1 answer