Question

A lot of 12 Compresser tanks is checked to see whether there are any defective tanks. three thanks are checked for leaks. if 1 or more of the 3 is defective, the lot is rejected. Find the probability that the lot will be rejected if three are actually 3 defective thanks in the lot.

Answer 1

X has the hypergeometric distribution:

$$𝑎 = 3 (𝑑𝑒𝑓𝑒𝑐𝑡𝑖𝑣𝑒), 𝑏 = 9 (𝑒𝑓𝑓𝑒𝑐𝑡𝑖𝑣𝑒)$$

$$𝑋 = 0 (𝑛𝑜𝑛𝑒 \ 𝑑𝑒𝑓𝑒𝑐𝑡𝑖𝑣𝑒)$$

$$n=3, 𝑛 − 𝑋 = 3$$

$$P(X)=\dfrac{\dbinom{a}{X}\dbinom{b}{n-X}}{\dbinom{a+b}{n}}$$

$$=\dfrac{\dbinom{3}{0}\dbinom{9}{3-0}}{\dbinom{3+9}{3}}$$

$$=\dfrac{1(84)}{220}=\dfrac{21}{55}$$

The probability that the lot will be rejected is

$$1-\dfrac{21}{55}=\dfrac{34}{55}\approx0.6182$$