A committee of 5 people is to be selected from student council. Council has 6 boys and 7 girls. What is the probability that the committee will have at least 3 boys?
*** If you can show work that would be helpful (OPTIONAL)***
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Define
C(n,r)=n!/(r!(n-r)!)=n choose r
Sample space: C(13,5)=1287
ways to choose 0 boy
=Choose 5 girls out of 7 and 0 boy out of 6
=C(7,5)*C(6,0)=21
ways to choose 1 boy:
=C(7,4)*C(6,1)=210
ways to choose 2 boys:
=C(7,3)*C(6,2)=525
ways to choose 3 boys:
=C(7,2)*C(6,3)=420
ways to choose 4 boys:
=C(7,1)*C(6,4)=105
ways to choose 5 boys:
=C(7,0)*C(6,5)=6
If you add them all up, they will total 1287 as required.
Now make appropriate sums according to requirements and calculate probability.
To find the probability that the committee will have at least 3 boys, we need to calculate the number of favorable outcomes (committees with at least 3 boys) and divide it by the total number of possible outcomes (all possible committees).
Step 1: Calculate the total number of possible committees:
Since we need to select a committee of 5 people, we can use the combination formula. The total number of possible committees can be calculated as C(13, 5), where 13 is the total number of students in the student council and 5 is the number of people in the committee.
C(13, 5) = 13! / (5!(13-5)!) = 1287
So, there are 1287 possible committees.
Step 2: Calculate the number of favorable outcomes (committees with at least 3 boys):
To calculate the number of committees with at least 3 boys, we need to consider two cases:
Case 1: Selecting exactly 3 boys and 2 girls.
Case 2: Selecting 4 boys and 1 girl.
Case 3: Selecting all 5 boys.
Case 1: Selecting exactly 3 boys and 2 girls:
We can choose 3 boys out of 6 (C(6, 3)) and 2 girls out of 7 (C(7, 2)).
C(6, 3) = 6! / (3!(6-3)!)= 20
C(7, 2) = 7! / (2!(7-2)!) = 21
The number of favorable outcomes for Case 1 = C(6, 3) * C(7, 2) = 20 * 21 = 420
Case 2: Selecting 4 boys and 1 girl:
We can choose 4 boys out of 6 (C(6, 4)) and 1 girl out of 7 (C(7, 1)).
C(6, 4) = 6! / (4!(6-4)!) = 15
C(7, 1) = 7! / (1!(7-1)!) = 7
The number of favorable outcomes for Case 2 = C(6, 4) * C(7, 1) = 15 * 7 = 105
Case 3: Selecting all 5 boys:
We can select all 5 boys out of 6 (C(6, 5)).
C(6, 5) = 6! / (5!(6-5)!) = 6
The number of favorable outcomes for Case 3 = C(6, 5) = 6
Total number of favorable outcomes = Sum of favorable outcomes for each case
= 420 + 105 + 6 = 531
Step 3: Calculate the probability:
Finally, we can calculate the probability by dividing the number of favorable outcomes by the total number of possible outcomes.
Probability = Number of favorable outcomes / Total number of possible outcomes
Probability = 531 / 1287 ≈ 0.4125
So, the probability that the committee will have at least 3 boys is approximately 0.4125.