your employer is offering free tickets for two basketball games. They play 53 home games and eight of them are against the snakes. The tickets are randomly given out. What is the probability that both games are against the snakes?
To find the probability that both games are against the snakes, we first need to determine the total number of possible outcomes.
There are a total of 53 home games and each game can be against any of the teams, including the snakes. Since 8 of the games are against the snakes, there are (53-8) = 45 games that are not against the snakes.
The total number of possible outcomes is then the number of ways to choose 2 games out of the total of 53 games, which is given by the combination formula:
C(n, r) = n! / (r! * (n-r)!)
where n is the total number of games (53) and r is the number of games we are choosing (2).
C(53, 2) = 53! / (2! * (53-2)!)
= 53! / (2! * 51!)
= (53*52) / (2*1)
= 1378
Next, we need to determine the number of favorable outcomes, which is the number of ways to choose 2 games out of the 8 games against the snakes:
C(8, 2) = 8! / (2! * (8-2)!)
= 8! / (2! * 6!)
= (8*7) / (2*1)
= 28
Therefore, the probability that both games are against the snakes is:
Favorable Outcomes / Total Outcomes
= 28 / 1378
= 0.0203
So, the probability that both games are against the snakes is approximately 0.0203 or 2.03%.