young_gun wrote:

John and Peter are among the nine players a basketball coach can choose from to field a five-player team. If all five players are chosen at random, what is the probability of choosing a team that includes John and Peter?

1/9

1/6

2/9

5/18

1/3

Total number of ways of choosing 5 people out of 9 = how many combinations are possible = 9C5.

Total number of favorite combinations, any of these combinations will always include John and Peter. thus we need to know how many combinations are feasible, when we have to choose 3 people out of 7 = 7C3.

Thus probability

P= 7C3/ 9C5 = 35/126 = 5/18.

Possible Wrong approach is as follows***********************
But initially, I had taken the wrong approach, and used the permutation fot the favorable events, I did something like this:

2 ( John and Peter) are already chosen, and now we need to choose the remaining 3 out of 7. Thus first can be chosen in 7 ways, second in 6 ways and third in 5 ways.

P= 7*6*5/ 9C5.

But 7*6*5 is permutation, just the number of ways the three people can be arranged and not the total number of combinations.

Just think/remember: ORDER does not matter in selecting the 3people out of 7, therefore it is a combination problem and not permutation ( in which order matters.)