Statistic Problem
1. Known the number of ball n(s) = 100
Ask: The change take a ball with number divisible by 5 but not divisible by 3 if taken randomly.
Answer: For example A = ball with number divisible by 5 but not divisible by 3 n(A) = 14, namely 5, 10, 20, 25, 35, 40, 50, 55, 65, 70, 80, 85, 95 and 100
P(A) = n(A)/n(s) = 14/100 = 7/50
2. Known: One badminton team consist of 5 members, will decide 2 people play a single players, and 2 pairs to play double.
Ask: How many option can be made.
First Way Solution:
For example 5 people of badminton team are A, B, C, D, and E. Then we choose A and B to play single.
The option number when A and B choose as single player are 6 option. The way to choose player as single player are 10, namely AB, AC, AD, AE, BC, BD, BE, CD, CE, and DE.
From one option of single player will make 6 option so for all way to choose single player will get 10 x 6 = 60 option.
Second way solution:
• The number option to choose 2 people as single player use the following formula:
C25 = 5!/(2!(5-2)!) = 5!/(2!.3!) = 10
• The number option to choose first pairs to play double
C23 = 3!/(2!(3-2)!) = 3!/(2!.1!) = 3
• The number option to choose second pairs to play double
C12 = 2!/(1!(2-1)!) = 2!/(1!.1!) = 2
So the number all option can be made are 10 x 3 x 2 = 60
Ask: The change take a ball with number divisible by 5 but not divisible by 3 if taken randomly.
Answer: For example A = ball with number divisible by 5 but not divisible by 3 n(A) = 14, namely 5, 10, 20, 25, 35, 40, 50, 55, 65, 70, 80, 85, 95 and 100
P(A) = n(A)/n(s) = 14/100 = 7/50
2. Known: One badminton team consist of 5 members, will decide 2 people play a single players, and 2 pairs to play double.
Ask: How many option can be made.
First Way Solution:
For example 5 people of badminton team are A, B, C, D, and E. Then we choose A and B to play single.
| Option | Single | Double |
1
| A, B | CD, AE |
2
| A, B | CE, AD |
3
| A, B | DE, AC |
4
| A, B | CD, BE |
5
| A, B | CE, BD |
6
| A, B | DE, BC |
The option number when A and B choose as single player are 6 option. The way to choose player as single player are 10, namely AB, AC, AD, AE, BC, BD, BE, CD, CE, and DE.
From one option of single player will make 6 option so for all way to choose single player will get 10 x 6 = 60 option.
Second way solution:
• The number option to choose 2 people as single player use the following formula:
C25 = 5!/(2!(5-2)!) = 5!/(2!.3!) = 10
• The number option to choose first pairs to play double
C23 = 3!/(2!(3-2)!) = 3!/(2!.1!) = 3
• The number option to choose second pairs to play double
C12 = 2!/(1!(2-1)!) = 2!/(1!.1!) = 2
So the number all option can be made are 10 x 3 x 2 = 60
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