Respuesta :
Answer:
The probability that no more than 3 of the entry forms will include an order is P=0.1243.
Step-by-step explanation:
We can model the probability of having k entries with order included within the 14 entries with a binomial distribution, with parameters n=14 and p=0.4.
The probabiltity of having k orders in the sample is:
[tex]P(x=k) = \dbinom{n}{k} p^{k}(1-p)^{n-k}\\\\\\P(x=k) = \dbinom{14}{0} 0.4^{k} 0.6^{14-k}\\\\\\[/tex]
We have to calculate the probability that no more than 3 of the entry forms will include an order. This is P(X≤3).
[tex]P(X\leq3)=P(x=0)+P(x=1)+P(X=2)+P(x=3)\\\\\\P(x=0) = \dbinom{14}{0} p^{0}(1-p)^{14}=1*1*0.0008=0.0008\\\\\\P(x=1) = \dbinom{14}{1} p^{1}(1-p)^{13}=14*0.4*0.0013=0.0073\\\\\\P(x=2) = \dbinom{14}{2} p^{2}(1-p)^{12}=91*0.16*0.0022=0.0317\\\\\\P(x=3) = \dbinom{14}{3} p^{3}(1-p)^{11}=364*0.064*0.0036=0.0845\\\\\\\\ P(X\leq3)=0.0008+0.0073+0.0317+0.0845=0.1243[/tex]
