Probability Distribution

Question

In a large town one person in five is left-handed.

(a) Find the probability that in a group of 30 person.

(i) exactly three will be left-handed

(ii) one third will be left handed

(b) How large must a sample be if the probability of at least one left handed person is to be greater than 90%.

I have done part (a), The problem i face in part be.

Answer 1

(1) The probability of choosing a left-handed person from the population in a large town is 1/5

P(left-handed) = 1/5

(2) The probability of choosing a right-handed person is 4/5

P(right-handed) = 4/5

(3) The probability that the group of 30 persons are all right-handed is (4/5)³⁰.

P(30 persons are all right-handed) = (4/5)³⁰

To get the probability that the group of 30 persons have at least one left handed is to subtract (3) from 100% (or 1).

P(at least 1 person is left-handed among the 30) = 1 - (4/5)³⁰

Therefore, we can say that:

P(at least 1 person is left-handed among the n-sample) = 1 - (4/5)ⁿ

(Always remember that n and 1-(4/5)ⁿ are directly proportional.)

Since we are looking for n-sample of greater than 90%, we can put the equation this way:

1 - (4/5)ⁿ > 0.90

Solve the inequality:

multiply both sides by -1 (reverse the symbol of inequality if multiplying or dividing both sides by a negative number):

-(4/5)ⁿ > -0.10

get the ln of both sides:

ln [(4/5)ⁿ] < ln (0.10)

Use the ln property ln (b^a) = a ln (b)

n• ln (4/5) < ln (0.10)

Divide both sides by ln (4/5), (note: ln of a number between 0 and 1 is a negative number):

n > ln (0.10)/ ln (4/5)

n > 10.31885...

Since we can't have fraction for the value of n, therefore the number of sample n should be at least 11 people.

Let's check:

1-(4/5)¹⁰ ≈ 0.8926

1- (4/5)¹¹ ≈ 0.9141

*There are only 89.26% for a sample of 10 and 91.41% for a sample of 11 and that is greater than 90%. Therefore our answer is correct.