Question

A class has 40 students. Each student is assigned a distinct number from 1 to 40. Four students will be selected at random. Find the probability of the event that the first student selected is assigned a number from 1 to 10, the second student selected is assigned a number from 11 to 20, the third student selected is assigned a number from 1 to 10, and the fourth student selected is assigned a number from 11 to 20 under the following sampling schemes:

a) random sampling with replacement
b) random sampling without replacement

Answer 1

a)

Solution

Four students will be selected randomly with replacement. So the choice doesn't affect the next choice. They are independent.

The probability the first selected student is assigned the number from 1 to 10 is $\frac{10}{40}$.

The probability the second selected student is assigned the number from 11 to 20 is again $\frac{10}{40}$, because there are still 40 students to select from.

The probability the third selected student is assigned the number from 1 to 10 is still $\frac{10}{40}$, because there are still 10 students who is assigned the number from 1 to 10.

The probability the fourth selected student is assigned the number from 11 to 20 is $\frac{10}{40}$.

All in all probability is $\frac{10}{40}\cdot\frac{10}{40}\cdot\frac{10}{40}\cdot\frac{10}{40} = \frac{1}{256} \approx 0.0039$

Answer: 0.0039

b)

Solution

Four students will be selected randomly without replacement. So the choice does affect the next choice.

The probability the first selected student is assigned the number from 1 to 10 is $\frac{10}{40}$. 39 students are left. 9 students who is assigned a number from 1 to 10 are left.

The probability the second selected student is assigned the number from 11 to 20 is $\frac{10}{39}$​. 38 students are left. 9 students who is assigned a number from 11 to 20 are left.

The probability the third selected student is assigned the number from 1 to 10 is $\frac{9}{38}$. 37 students are left.

The probability the fourth selected student is assigned the number from 11 to 20 is $\frac{9}{37}$.

All in all probability is $\frac{10}{40}\cdot\frac{10}{39}\cdot\frac{9}{38}\cdot\frac{9}{37} \approx 0.00369$

Answer: 0.00369