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
2) one student is absent
3) three student is absent
4) at least one student is absent
To find the probability of each event, we will use the binomial probability formula:
P(X=k) = (nCk) * (p^k) * ((1-p)^(n-k))
Where:
P(X=k) is the probability of having exactly k successes,
n is the number of trials,
p is the probability of success in a single trial,
k is the number of successful events.
1) To find the probability that no student is absent, we have:
n = 20 (number of students)
p = 0.09 (probability of each student being absent)
k = 0 (number of students absent)
P(X=0) = (20C0) * (0.09^0) * ((1-0.09)^(20-0))
P(X=0) = (1) * (1) * (0.91^20)
P(X=0) ≈ 0.305
Therefore, the probability that no student is absent is approximately 0.305.
2) To find the probability that one student is absent, we have:
n = 20 (number of students)
p = 0.09 (probability of each student being absent)
k = 1 (number of students absent)
P(X=1) = (20C1) * (0.09^1) * ((1-0.09)^(20-1))
P(X=1) = (20) * (0.09) * (0.91^19)
P(X=1) ≈ 0.382
Therefore, the probability that one student is absent is approximately 0.382.
3) To find the probability that three students are absent, we have:
n = 20 (number of students)
p = 0.09 (probability of each student being absent)
k = 3 (number of students absent)
P(X=3) = (20C3) * (0.09^3) * ((1-0.09)^(20-3))
P(X=3) = (1140) * (0.09^3) * (0.91^17)
P(X=3) ≈ 0.319
Therefore, the probability that three students are absent is approximately 0.319.
4) To find the probability that at least one student is absent, we can find the complement of the probability that no student is absent.
P(at least one student absent) = 1 - P(X=0)
P(at least one student absent) = 1 - 0.305
P(at least one student absent) ≈ 0.695
Therefore, the probability that at least one student is absent is approximately 0.695.
To find the probability of each scenario, we need to use the concept of independent events and the probability of individual events occurring. In this case, the probability of a student missing class is given as 9% or 0.09.
1) To find the probability that no student is absent, we calculate the probability of each student attending class. Since attendance is independent, we multiply the probabilities together.
P(no student absent) = (1 - 0.09)^20 = 0.91^20 ≈ 0.1216
So, the probability that no student is absent is approximately 0.1216.
2) To find the probability that exactly one student is absent, we need to consider the different ways this can happen. There are 20 students in the class, so any one student can be absent. We then multiply this by the probability of one student being absent and the rest attending.
P(one student absent) = 20 * (0.09) * (0.91)^19 ≈ 0.2707
So, the probability that exactly one student is absent is approximately 0.2707.
3) To find the probability that exactly three students are absent, we use the same logic. We need to consider the different combinations of three students being absent, multiplied by the probability of those three being absent and the rest attending.
P(three students absent) = (20 choose 3) * (0.09)^3 * (0.91)^17 ≈ 0.2196
So, the probability that exactly three students are absent is approximately 0.2196.
4) To find the probability that at least one student is absent, we can find the complement probability, which is the probability that no student is absent, and then subtract it from 1.
P(at least one student absent) = 1 - P(no student absent) ≈ 1 - 0.1216 ≈ 0.8784
So, the probability that at least one student is absent is approximately 0.8784.