# Expected Value

• Sep 24th 2009, 07:43 PM
chengbin
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 $k \cdot ...$?
• Sep 24th 2009, 11:19 PM
mr fantastic
Quote:

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.
• Sep 25th 2009, 04:09 AM
chengbin
Why is it $k \cdot P_k$?
• Sep 25th 2009, 04:14 AM
mr fantastic
Quote:

Originally Posted by chengbin
Why is it $k \cdot P_k$?

This follows directly from the definition of expected value.