3:A basketball game in a league between the sharks and the jets was scheduled for thursday night. Only four players from the sharks showed up and so the jets, who had six players won by default, so the evening would not be spoiled ,the players arrieved four the game decided to have a pickup scrimmage and picked the two teams by drawing names from a hat,what is the probability that rhe four sharks end up on the same team?
I will be happy to critique your thinking.
To determine the probability that the four Sharks players end up on the same team during the pickup scrimmage, we need to calculate the number of favorable outcomes (where all four Sharks players are on the same team) and divide it by the total number of possible outcomes.
There are a total of 10 players for the pickup scrimmage (4 Sharks players + 6 Jets players). Since the players are being selected randomly by drawing names from a hat, we can consider the problem as drawing without replacement.
First, let's calculate the total number of possible outcomes. We need to choose 5 players from the hat (since 10 players are available) to form one team. This can be done using the combination formula:
C(10, 5) = 10! / (5! * (10-5)!) = 252
Now, let's calculate the number of favorable outcomes, where all four Sharks players end up on the same team. We need to choose 4 players from the remaining 9 (since one Sharks player has already been selected for the team) to join the Sharks players. Again, this can be done using the combination formula:
C(9, 4) = 9! / (4! * (9-4)!) = 126
Therefore, the probability that the four Sharks players end up on the same team is:
P(Favorable) = Number of favorable outcomes / Total number of possible outcomes
P(Favorable) = 126 / 252 = 0.5
So, the probability that the four Sharks players end up on the same team is 0.5 or 50%.