SOLUTION: 8% of students are left handed. If 50 (FIFTY) classes of 20 (TWENTY) students are randomly selected, what is the probability that 10 (TEN) classes have no left-handed students?

Question 1023484: 8% of students are left handed. If 50 (FIFTY) classes of 20 (TWENTY) students are randomly selected, what is the probability that 10
(TEN) classes have no left-handed students?
Answer by mathmate(429) (Show Source): You can put this solution on YOUR website!

Question:8% of students are left handed. If 50 (FIFTY) classes of 20 (TWENTY) students are randomly selected, what is the probability that 10
(TEN) classes have no left-handed students?

Solution:
This question can be broken down into two steps.
The first step is to apply the multiplication rule to find the probability of a class of 20 right-handed students.
We need to assume that the students (and classes) are chosen randomly and independent of each other.
Probability of success, i.e. 20 right-handed students, is obtained from the probability of randomly choosing a single right-handed student, equal to p=1-0.08=0.92. probability of a right-handed class will therefore be p^20, by the multiplication rule, where p^20=0.92^20=0.18869.

Now we have 50 classes, either right-handed or mixed. To find the probability of (exactly) 10 classes out of the 50 are right-handed can be obtained by the binomial distribution, namely
P(k;n;p)=C(n,k)*p^k*(1-p)^(n-k)
where C(n,k) is the binomial coefficient = n!/(k!(n-k)!)
p=probability of success (right-handed class)
(1-p)=probability of failure (mixed class), and
k=number of successes.

Applying the formula, we have
P(10;50;0.18869)=C(50,10)*0.18869^10*(0.81131)^40
=10272278170*5.72223*10^(-8)*2.33025*10^(-4)
=0.13697
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