Answer: 0.30
Step-by-step explanation:
Binomial distribution formula :-
[tex]P(x)=^nC_xp^x(1-p)^{n-x}[/tex], where P(x) is the probability of getting success in x trials , n is the total number of trials and p is the probability of getting success in each trial.
Given : The probability that a shipment are known to be defective= 0.07
If a sample of 5 items is randomly selected from this shipment,then the probability that at least one defective item will be observed in this sample will be :-
[tex]P(X\geq1)=1-P(0)\\\\=1-(^5C_0(0.07)^0(1-0.07)^{5-0})\\\\=1-(0.93)^5=0.3043116307\approx0.30[/tex]
Hence, the probability that at least one defective item will be observed in this sample =0.30
