# Thread:

1. ## Expected Value

A sack contains 20 products, where two are defective. Given that five products are chosen randomly, determine the expected number of defective products.

Let $P_k$ be the probability of there being k defective products out of the 5 chosen.

$P_k=\frac {_2C_k \cdot _{18}C_{5-k}}{_{20}C_5}$

$\therefore E(X) = 0 \cdot \frac {_2C_0 \cdot _{18}C_5}{_{20}C_5} + 1 \cdot \frac {_2C_1 \cdot _{18}C_4}{_{20}C_5} + 2 \cdot \frac {_2C_2 \cdot _{18}C_3}{_{20}C_5}$

Why is it $k \cdot ...$?

2. Originally Posted by chengbin
A sack contains 20 products, where two are defective. Given that five products are chosen randomly, determine the expected number of defective products.

Let $P_k$ be the probability of there being k defective products out of the 5 chosen.

$P_k=\frac {_2C_k \cdot _{18}C_{5-k}}{_{20}C_5}$

$\therefore E(X) = 0 \cdot \frac {_2C_0 \cdot _{18}C_5}{_{20}C_5} + 1 \cdot \frac {_2C_1 \cdot _{18}C_4}{_{20}C_5} + 2 \cdot \frac {_2C_2 \cdot _{18}C_3}{_{20}C_5}$

Why is it $k \cdot ...$?
I don't really understand the trouble here .... k is the number of defectives in the chosen sample.

3. Why is it $k \cdot P_k$?

4. Originally Posted by chengbin
Why is it $k \cdot P_k$?
This follows directly from the definition of expected value.
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