Suppose that each student in stat. has a 9% chance of missing class on any given day and that student attendance is independent. In a stat. class of 20 students, find the probability that:
1) no student is absent =15.16
2) one student is absent =30
3) three student is absent =16.72
4) at least one student is absent =31
Try a binomial probability table or use a binomial probability function, which is this:
P(x) = (nCx)(p^x)[q^(n-x)]
For 1), find P(0):
x = 0
n = 20
p = .09
q = .91 (q is 1-p)
For 2), find P(1)
x = 1
n,p,q same as 1)
For 3), find P(3)
x = 3
n,p,q same as 1)
For 4), take 1 - P(0).
I'll let you take it from here.
To find the probability in each case, we can use the binomial probability formula, given by:
P(x) = (nCx) * p^x * (1-p)^(n-x)
Where:
P(x) is the probability of exactly x successes (students absent) in a given scenario
n is the total number of trials (number of students in the class)
x is the number of successes (number of students absent)
p is the probability of a success on any given trial (probability of a student missing class)
(1-p) is the probability of a failure on any given trial (probability of a student attending class)
Now, we can calculate the probabilities for each scenario:
1) No student is absent:
P(0) = (20C0) * (0.09^0) * (0.91^(20-0))
= 1 * 1 * 0.1216
= 0.1216
So, the probability that no student is absent is 0.1216 or 12.16%.
2) One student is absent:
P(1) = (20C1) * (0.09^1) * (0.91^(20-1))
= 20 * 0.09 * 0.1847
= 0.33246
So, the probability that one student is absent is 0.33246 or 33.25%.
3) Three students are absent:
P(3) = (20C3) * (0.09^3) * (0.91^(20-3))
= (1140) * (0.000729) * (0.1963)
= 0.13416532
So, the probability that three students are absent is 0.13416532 or 13.42%.
4) At least one student is absent:
To find the probability that at least one student is absent, we can subtract the probability of no students being absent from 1.
P(at least one) = 1 - P(0)
= 1 - 0.1216
= 0.8784
So, the probability that at least one student is absent is 0.8784 or 87.84%.