There’s a competition with 6 people. 3 of the 6 advance to the next round. 3 of the 6 are representing the same team. what’s the percentage chance of the previously mentioned team advancing at least one person?

The percentage of at least one person of the team advanced is 95%

Step-by-step explanation:

For this exercise is necessary to understand the concept of combination, this is calculate as:

Where n is the total elements and k is the size of the group that is going to be chosen. This calculation give as the number of ways that is possible to choose a group of k with n elements.

So, the only possibility that there is no one of the same team that advance to the next round is that all of the people chosen were of the other 3 that doesn’t belong to the group mentioned. Then there is only one way to get this case.

On the other hand there are 20 ways to conform a group of 3 from a 6 elements and is calculate as:

Therefore, the percentage that any of the group mentioned pass to the next round is 1 in 20 or 5%.  

So, the percentage that at least one of the group mentioned pass to the next round is the compliment, that’s mean that it is 95%.

4th option is correct xy=-10

if this was to u plzz mark me the brainliest

a

step-by-step explanation:

the answer is a