Three members of a 12 person singing group are going to be chosen to sing a song in their coming concert.
A) how many different groups containing 3 people can be chosen from the 12 members of the singing group?
B) what Is the probability that mike and Jo will both be in the selected group?
b. three choices
?MJ
?JM
J?M
JM?
MJ?
M?J so there are six ways to get it, so probability is 6*(1/12*1/11*10/10)
a. 3!/12!9!
oops a. 12!/(3!9!)=12*11*10/6=11*10*2
check that
A) To find the number of different groups containing 3 people that can be chosen from the 12 members of the singing group, we can use the concept of combinations. In this case, we want to find the number of combinations of 12 members taken 3 at a time.
The formula for combinations is given by:
C(n, r) = n! / (r!(n-r)!)
Here, n represents the total number of members (12), and r represents the number of members to be chosen for each group (3).
Plugging in the values into the formula:
C(12, 3) = 12! / (3!(12-3)!)
= 12! / (3!9!)
12! means factorial of 12, which is calculated as:
12! = 12 × 11 × 10 × 9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1
3! means factorial of 3, which is calculated as:
3! = 3 × 2 × 1
9! means factorial of 9, which is calculated as:
9! = 9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1
Simplifying the expression:
C(12, 3) = (12 × 11 × 10 × 9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1) / ((3 × 2 × 1)(9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1))
Calculating the numerator and denominator:
C(12, 3) = (12 × 11 × 10) / (3 × 2 × 1)
= (1320) / (6)
= 220
Therefore, there are 220 different groups containing 3 people that can be chosen from the 12 members of the singing group.
B) To find the probability that both Mike and Jo will be in the selected group, we need to calculate the number of groups containing Mike and Jo divided by the total number of groups that can be formed.
The number of groups containing Mike and Jo can be calculated as the number of combinations of 10 remaining members taken 1 at a time (since 2 members are already fixed). Using the same formula as above:
C(10, 1) = 10! / (1!(10-1)!)
= 10! / (1!9!)
Simplifying the expression:
C(10, 1) = (10 × 9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1) / (1 × (9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1))
= 10
Therefore, there are 10 different groups containing Mike and Jo.
The total number of groups that can be formed is given by the answer to part A, which is 220.
To find the probability, we divide the number of groups containing Mike and Jo (10) by the total number of groups (220):
Probability = 10 / 220
= 1 / 22
So, the probability that Mike and Jo will both be in the selected group is 1/22.