in a certain college 33% of the physics majors belong to ethnic minorities. if 8 students are selected at random from the physics majors what is the probability that more than 5 belong to an ethnic minority
To find the probability that more than 5 students belong to an ethnic minority, we need to calculate the probability of selecting exactly 6, 7, or 8 physics majors who belong to an ethnic minority.
First, let's calculate the probability of selecting two different groups: non-minority students and minority students.
1. Probability of selecting a non-minority student: 1 - 0.33 = 0.67
2. Probability of selecting a minority student: 0.33
To calculate the probability of each case, we'll use the binomial probability formula:
P(x) = (nCx) * (p^x) * (q^(n-x))
Where:
- n is the total number of trials (8 students selected)
- x is the number of successful outcomes (≥ 6 students from an ethnic minority)
- p is the probability of a successful outcome (probability of selecting a minority student)
- q is the probability of an unsuccessful outcome (probability of selecting a non-minority student)
Now, let's calculate the probabilities for each case:
1. Selecting exactly 6 students from an ethnic minority:
P(6) = (8C6) * (0.33^6) * (0.67^2)
2. Selecting exactly 7 students from an ethnic minority:
P(7) = (8C7) * (0.33^7) * (0.67^1)
3. Selecting exactly 8 students from an ethnic minority:
P(8) = (8C8) * (0.33^8) * (0.67^0)
Finally, we'll sum up the probabilities for these three cases to get the probability of more than 5 students belonging to an ethnic minority:
P(more than 5) = P(6) + P(7) + P(8)
To find the probability that more than 5 students belong to an ethnic minority out of a random selection of 8 students from the physics majors, we'll need to use the binomial distribution formula.
The binomial distribution formula is given by:
P(X = k) = C(n, k) * p^k * (1 - p)^(n - k)
Where:
P(X = k) represents the probability of getting k successes in n trials,
C(n, k) is the combination function, used to calculate the number of ways to choose k successes from n trials,
p is the probability of a single success, and
(1 - p) is the probability of a single failure.
In this case, the probability of a single success is the proportion of physics majors belonging to ethnic minorities, which is 33% or 0.33, and the probability of a single failure is the remaining proportion, which is 1 - 0.33 = 0.67.
To solve this problem, we need to calculate the following probabilities:
P(X = 6) + P(X = 7) + P(X = 8), which represents the probability of getting 6, 7, or 8 students who belong to an ethnic minority out of the 8 randomly selected students.
Let's calculate each of these probabilities step by step:
P(X = 6) = C(8, 6) * 0.33^6 * 0.67^2
P(X = 7) = C(8, 7) * 0.33^7 * 0.67^1
P(X = 8) = C(8, 8) * 0.33^8 * 0.67^0
To calculate the combinations in the formula, use the combination formula:
C(n, k) = n! / (k! * (n - k)!)
Let's calculate:
We need the sum of probabilities of 6, 7 and 8
This is much more advanced than your previous questions and requires use of the binomial distribution rules.
P(x=k) = C(n,k) p^k p^(n-k)
where C(n,k)=n!/[k!(n-k)!]
so for example the P(x=6) is
P(x=6)= C(8,6) .33^6 * .67^2
where C(8,6) = 8!/[6!*2!]=28
so
P(x=6) = 28* .33^6 * .67^2
= 0.01623
now do that for P(x=7)
and for P(x=8)
and add the three results.