Given:
The number of players on the basketball = 14 players.
The number of juniors =6 players.
The number of seniors = 8 players.
Coach Banet decided to choose three players.
Required:
A. We need to find different orders of the top three finishers.
B. We need to find the probability that the top three finishers will all be seniors.
Explanation:
A.
There is not important to choose in the order which players are the top three finishers.
Use combinations.
The number of students, n=14.
The number of the top three finishers, r =3.
[tex]nC_r=14C_3[/tex]
[tex]=\frac{14!}{3!(14-3)!}[/tex]
[tex]=\frac{14!}{3!\cdot11!}[/tex]
[tex]=364[/tex]
Answer:
[tex]nC_r=\frac{14!}{3!\cdot11!}=364[/tex]
B.
The number of seniors = 8 players.
The number of top-finishers =3.
[tex]8C_3=\frac{8!}{3!(8-3)!}=\frac{8!}{3!\cdot5!}=56[/tex]
There are 56 different orders of top finishers that include all seniors.
[tex]The\text{ possible outcome=}14C_3[/tex][tex]T\text{he favorable outcome =}8C_3[/tex]
The probability that the top three finishers will all be seniors.
[tex]P=\frac{8C_3}{14C_3}[/tex][tex]Use\text{ }14C_3=364,\text{ and }8C_3=56.[/tex]
[tex]P=\frac{56}{364}[/tex]
[tex]P=0.1538[/tex]
Multiply by 100 to get a percentage.
[tex]P\text{ \%}=0.1538\times100[/tex]
[tex]P\text{ \%}=15.38\text{ \%}[/tex]
[tex]P\text{ \%}=15.4\text{ \%}[/tex]
Answer:
There are 56 different orders of top finishers that include all seniors.
The probability that the top three finishers will all be seniors is 15.4 %
1)
B)
The number of juniors = 6 players.
The number of players in the group =3.
There is not important to choose in the order which players are selecting.
Use combinations.
[tex]6C_3=\frac{6!}{3!(6-3)!}=\frac{6!}{3!\cdot3!}=20[/tex]
Answer:
Two terms represent the number of players that are all juniors.
[tex]6C_3\text{ }and\text{ }20[/tex]
