Question 1042072: Five students will be randomly selected from the seven top students in mathematics to represent their school in a regional competition. Angela (A) and Brian (B) are among the seven students, and they wish to determine the probability that they will both be chosen. Using A, B, C, D, E, F, and G to denote the students, set up a sample space to determine this probability. What is the probability they will both be chosen? Enter probability as a fraction.
Answer by jim_thompson5910(35256)   (Show Source):
You can put this solution on YOUR website!
Use the combination formula to get
n C r = (n!)/(r!(n-r)!)
7 C 5 = (7!)/(5!*(7-5)!)
7 C 5 = (7!)/(5!*2!)
7 C 5 = (7*6*5!)/(5!*2!)
7 C 5 = (7*6)/(2!)
7 C 5 = (7*6)/(2*1)
7 C 5 = (42)/(2)
7 C 5 = 21
So there are 21 ways to pick 5 students. The sample space would be...
A,B,C,D,E
A,B,C,D,F
A,B,C,D,G
A,B,C,E,F
A,B,C,E,G
A,B,C,F,G
A,B,D,E,F
A,B,D,E,G
A,B,D,F,G
A,B,E,F,G
A,C,D,E,F
A,C,D,E,G
A,C,D,F,G
A,C,E,F,G
A,D,E,F,G
B,C,D,E,F
B,C,D,E,G
B,C,D,F,G
B,C,E,F,G
B,D,E,F,G
C,D,E,F,G
If you counted all the possibilities in the sample space above, you'd count out 21 different items. Each item has its own line.
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Now let's lock up the first two slots for students A and B. That leaves 7-2 = 5 students left and 5-2 = 3 slots left
So using the combination formula again gives
n C r = (n!)/(r!(n-r)!)
5 C 3 = (5!)/(3!*(5-3)!)
5 C 3 = (5!)/(3!*2!)
5 C 3 = (5*4*3!)/(3!*2!)
5 C 3 = (5*4)/(2!)
5 C 3 = (5*4)/(2*1)
5 C 3 = (20)/(2)
5 C 3 = 10
There are 10 ways to pick a group of five students (from the pool of seven) where students A and B are guaranteed to go.
The 10 different ways are listed below
A,B,C,D,E
A,B,C,D,F
A,B,C,D,G
A,B,C,E,F
A,B,C,E,G
A,B,C,F,G
A,B,D,E,F
A,B,D,E,G
A,B,D,F,G
A,B,E,F,G
That list of 10 is a subset of the first list of items shown above.
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Now divide the two values to get 10/21 which is the final answer
Side note: 10/21 = 0.47619047619048 = 47.619047619048%
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Final Answer: 10/21
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