A simple random sample of 5 students is selected, without replacement, from a gr

Question

A simple random sample of 5 students is selected, without replacement, from a group of 50 UL Lafayette students. If 80% of the students in this group were born in Louisiana, find the probability of the following events.

a) All five of the students were born in Louisiana.

b) The first student chosen for the sample and the fifth student chosen for the sample were not born in Louisiana but the other three were.

c) Two of the five students were not born in Louisiana but the other three were.

d) The third student selected for the sample was born in Louisiana.

The binomial distribution with parameters n (number of trials and p (success probability). Restrictions on the parameters: n is a positive integer (1,2,3,.. and 0

Explanation / Answer

Total number of students = 50

Number of students born in Louisiana = 0.8*50 = 40

a.) All five born in Louisisna = (40/50)*(39/49)*(38/48)*(37/47)*(36/46) = 0.3105

b.) P = (10/50)*(40/49)*(39/48)*(38/47)*(9/46) = 0.0209

c.) P = (10/50)*(40/49)*(39/48)*(38/47)*(9/46)* (5!/(2!*3!) = 0.209

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