Answer:
Step-by-step explanation:
P\left ( defective item\right )=0.035
Using binomial distribution
Where p= probability of success
q=probability of failure
Here p=0.035
q=1-0.035=0.965
[tex]^nC_{r}P^{r}q^{n-r}[/tex]
(i)for exactly 3 defective items i.e. r=3
P[tex]\left ( r=3\right )[/tex]=[tex]^{44}C_{3}[/tex][tex]\left ( 0.035\right )^{3}\left ( 0.965\right )^{44-3}[/tex]
P=[tex]\frac{44!}{41!3!}\times \left ( 0.035\right )^3\left ( 0.965\right )^{41}[/tex]
P=0.1317
(ii)No defective item i.e. r=0
P[tex]\left ( r=0\right )[/tex]=[tex]^{44}C_{0}[/tex][tex]\left ( 0.035\right )^{0}[/tex][tex]\left ( 0.965\right )^{44-0}[/tex]
P=[tex]\frac{44!}{44!0!}\times \left ( 0.035\right )^0\left ( 0.965\right )^{44}[/tex]
P=0.2085
(iii)At least 1 defective item
P=1-P(zero defective item)
P=1-[tex]^{44}C_{1}\left ( 0.035\right )^{1}\left ( 0.965\right )^{44-1}[/tex]
P=1-[tex]\frac{44!}{43!1!}\times \left ( 0.035\right )^1[/tex][tex]\left ( 0.965\right )^{43}[/tex]
P=0.6671