The expected number of distinct units in a simple random sample of 3 units drawn with replacement from a population of 100 units is  

Concept:

Expected Number of Distinct Units:

The expected number of distinct units can be calculated using the complement rule,

which states that the expected number of distinct units in a sample of size n from a population

of N can be found by subtracting the probability of repeated selections from 1.

The general formula for the expected number of distinct elements in a sample of size k

 from a population of N when sampling with replacement is:

$E(n)=N(1-{\left(\frac{N-1}{N}\right)}^k)$

This formula breaks down as

N: Total population size (i.e., 100 in this case).

$(1-{\left(\frac{N-1}{N}\right)}^k)$ represents the probability that not all selected units are the same, leading to distinct units.
   
Explanation:

Population size  N = 100 , sample size  n = 3

The probability of selecting a particular unit and not getting it again when sampling with

replacement is $\left(\frac{99}{100}\right)$. This is because there are 99 other units that can be chosen in the subsequent draws.

The formula to calculate the expected number of distinct units is based on the complement

of selecting the same unit across the draws. The general formula for the expected number of distinct

units, $E(k)$, in a sample of size $k$ from a population of N with replacement is:

 $E(k)=N(1-{\left(\frac{N-1}{N}\right)}^k)$

For this case, with N = 100 and k =3, the expected number of distinct units becomes:

  $E(3)=100(1-{\left(\frac{99}{100}\right)}^3)$

First, calculate the probability:

 ${\left(\frac{99}{100}\right)}^3=\frac{{99}^3}{{100}^3}=\frac{970299}{1000000}$
 

Now, subtract it from 1:

  $1-\frac{970299}{1000000}=\frac{1000000-970299}{1000000}=\frac{29701}{1000000}$

 Multiply by 100:

 $E(3)=100\times \frac{29701}{1000000}=\frac{29701}{10000}=2.9701$
 
The correct expression in the provided options that matches this logic is option 2).