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The student population of a certain college has been found to be made up as foll

ID: 3232492 • Letter: T

Question

The student population of a certain college has been found to be made up as follows: 40% are in-state students, 20% are out-of -state students from other states in the U.S. other than the New England states, 10% are from the New England states and 30% are foreign students. If 20 students from the college are randomly selected, determine the following:

a) The probability that 10 of them are in-state students, 5 are non-New England out -of-state students, 3 of them are from the New England states and 2 are foreign students.

b) The probability that 10 of them are in-states students and 5 of them are from New England states.

c)The probability that 10 of them are in-state students. Please give answers in written detail. Thanks very much.

Explanation / Answer

40% are in state

20% are out of state other than the New England states

10% are from New England state

30% are foreign students

Let assume total students be 100 (can be any number or even x)

Then

40 are from state

20 are out of state other than the New England states

10 are from New England state

30 are foreign students

20 students are randomly selected

Total ways of selecting 20 students out of 100 = 100C20 = 535983370403809682970 (here 100C20 is the number of ways 20 students can be selected from 100 students)

535983370403809682970 / (40! * 10! * 20! * 30!)as the same should be counted ones

a)

The probability that 10 of them are in-state students, 5 are non-New England out -of-state students, 3 of them are from the New England states and 2 are foreign students

     = 40C10 * 20C5 * 10C3 * 30C2

     = 847660528 * 15504 * 120 * 435

Probability = (847660528 * 15504 * 120 * 435 ) / 535983370403809682970

                 = 0.00127992

b) probability that 10 of them are in-states students and 5 of them are from New England states

    = 40C10 * 10C5 * 50C5(remaing 5 students from the remaing place)

= 847660528 * 252 * 2118760

probability = (847660528 * 252 * 2118760) / 535983370403809682970

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