Question

a starting lineup in basketball consist of a left guard, a right guard, a left forward, a right forward, and a center. Also, a certain regional high school serves students from the municipalities of Alicetown and Burridge, and this high school has a basketball team. Of the team's 4 guards, 2 are from Alicetown; of the team's 5 forwards, 2 are from Alicetown; and, of the team's 5 centers, 2 are from Alicetown.

a) How many ways are there to form a starting lineup consisting only of students from alicetown?

b) what is the probability that all the students in the starting lineup come from Alicetown?

c) What is the probability that all of the students in a starting lineup come from the same municipality?

Answer 1

a)Starting lineup consist of 5 students of which 2 are from guard and 2 are from forward and 1 from centre.

So, one can select 2 guard from team's 4 card in (4c2) ways

And, 2 forward from team's 5 forward in (5c2) ways

And , 1 centre students can be selected from team's 5 centre in (5c1) ways.

Therefore, total number of ways of selecting students from only alicetown in starting lineup=[(4c2)*(5c2)*(5c2)]= 6*10*10=600 ways.

b) total number of ways of selecting starting lineup student (i.e. 5 students) from (4+5+5)=14 students in 14c5 ways.

So, Probability that all students from starting lineup comes from Alicetown= 600/14c5= 600/2002=0.2997

c) since, the starting lineup consist of 5 students therefore, only team's 5 forward and team's 5 centre will able to make starting lineup individually and team's 4 guard can't make only by itself because it has only 4 students.

So, probability that all students in starting lineup came from same municipality= P( either starting lineup made by team's 5 forward or team's 5 centre)= (5c5/14c5)+(5c5/14c5)= (1/2002)+(1/2002)=2/2002=1/1001=0.00099