There are 6 tight ends in the Hall of Fame. How many different combinations of 2 of those players could possibly be selected for an all-time team? What is the probability that a random selection of 2 tigh ends would be the same 2 chosen for the team?
I think there are 15 combinations.
Can I have a websie on probability?
That was supposed to say: Can some one give me a website on probability?
go on google type in probability and click on the 2nd search.
"15 combinations" is correct.
Since there are 15 combinations, the probability is 1/15.
To find the number of different combinations of 2 players that can be selected from 6 tight ends, we can use the concept of combinations in combinatorics.
The number of combinations of selecting 2 players from a set of 6, which can be represented as "6 choose 2" or written as C(6, 2), can be calculated as:
C(6, 2) = 6! / (2! * (6-2)!) = 6! / (2! * 4!) = (6 * 5) / (2 * 1) = 15
Therefore, there are 15 different combinations of 2 players that could possibly be selected for an all-time team.
To find the probability that a random selection of 2 tight ends would be the same 2 chosen for the team, we need to know the total number of possible outcomes.
The total number of possible outcomes would be the number of combinations of selecting 2 players from the total pool of 6 tight ends, which we calculated as 15.
Now, since we're interested in one specific outcome (selecting the same 2 players for the team), there is only 1 favorable outcome. Therefore, the probability of randomly selecting the same 2 players for the team is:
Probability = Favorable outcomes / Total outcomes
Probability = 1 / 15
Hence, the probability that a random selection of 2 tight ends would be the same 2 chosen for the team is 1/15, or approximately 0.067, which can also be expressed as a percentage of about 6.7%.