Question

During the manufacture of a product, the probability that the product will be satisfactory is 90%. The products are sampled individually from a conveyor belt. Out of 5 samples, what is the probability of finding exactly 2 defective products? Use the binomial distribution table and give the exact value. Hint: A "success" in this case means a defective product.

Answer 1

$\dpi{100} \mathbf{Solution}$

$\dpi{100} \mathbf{given:}\;n=\mathbf{5},\;p\;(defective)=1-P(satisfactory)=1-0.90=\mathbf{0.10},$

$\dpi{100} q(satisfactory)=\mathbf{0.90}$

$\dpi{100} \mathbf{formula:}\;P(X=r)=_{r}^{n}\textrm{C}*p^{r}*q^{n-r}\;\;\;where,\;\;_{r}^{n}\textrm{C}=\frac{n!}{r!*(n-r)!}$

$\dpi{100} \mathbf{\mathbf{P({\color{Blue} exactly}\;2\;defective\; products)}\Rightarrow P(X{\color{Blue} =}2)}$

$\dpi{100} P(X=2)=_{2}^{5}\textrm{C}*0.10^{2}*0.90^{5-2}$

$\dpi{100} P(X=2)=\frac{5!}{2!*(5-2)!}*0.10^{2}*0.90^{3}$

$\dpi{100} P(X=2)=\frac{5!}{2!*3!}*0.10^{2}*0.90^{3}$

$\dpi{100} P(X=2)=\frac{5*4*{\color{Blue} 3!}}{(2*1)*{\color{Blue} 3!}}*0.10^{2}*0.90^{3}$

$\dpi{100} P(X=2)=10*0.10^{2}*0.90^{3}$

$\dpi{100} P(X=2)=\mathbf{{\color{Red} 0.0729}}$