Ask your own question, for FREE!
Mathematics 15 Online
OpenStudy (anonymous):

Dave is the manager of a construction supply warehouse and notes that 60% of the items purchases are heating items, and 15 % are plumbing items. Find the probablility that at least three out of the next five items purchased are heating items. Some one please help its urgents, and oh Happy New Years!

OpenStudy (anonymous):

at least 3, this is not going to be that quick

OpenStudy (anonymous):

Its alright I just really dont understand this question, ill pay attention though

OpenStudy (anonymous):

you have to compute exactly 3, exactly 4, exactly 5 and exactly 6 and add them up OR you have to compute exactly 0, exactly 1 and exactly 2, add them up and subtract the total from 1

OpenStudy (anonymous):

oh three out of FIVE i should learn to read, in any case you will have to compute three probabilities

OpenStudy (anonymous):

the probability that all 5 are plumbing items is the easiest, it is \[P(x=5)=(.15)^5\] that is five times in a row you got plumbing items

OpenStudy (anonymous):

the probability that you get exactly 4 plumbing items is \[P(x=4)=4\times (.15)^4\times (.85)\]

OpenStudy (anonymous):

that is, you get exactly 4 plumbing items, exactly 1 not plumbing item, and there are 5 different ways to get that damn i made a typo sorry let me fix is replace the 4 by a 5

OpenStudy (anonymous):

\[P(x=4)=5\times (.15)^4\times (.85)\]

OpenStudy (anonymous):

finally the probability you get exactly 3 is \[P(x=3)=10\times( .15)^3\times (.85)^2\]

OpenStudy (anonymous):

compute these three numbers, add them up, and that will give you the probability that you get at least 3

OpenStudy (anonymous):

ok I will, but its the set up your using a formula? Because i remember my teacher using a long formula

OpenStudy (anonymous):

the formula for this binomial probability is this if you have \(n\) trails, with the probability of success in any one trial as \(p\) and therefore the probability of failure is therefore \(1-p\) then the probability you get exactly \(k\) successes in \(n\) trials is \[P(x=k)=\dbinom{n}{k}p^k(1-p)^{n-k}\]

OpenStudy (anonymous):

in your case we used \(n=5\), \(p=.15\) and \(1-p=.85\) but we had to compute for \(k=3, k=4,k=5\)

OpenStudy (anonymous):

note that \(\binom{5}{3}=10\) and \(\binom{5}{4}=5\)

OpenStudy (sirm3d):

\(p=0.6\) for heating items, \(q=0.4\) for non-heating items

OpenStudy (anonymous):

when i do .15^5 i get a weird number on my calculator? DO you have the same?

OpenStudy (anonymous):

so it is my answers were for plumbing because i can't read

OpenStudy (anonymous):

OH NO!!! Lol we screwed up

OpenStudy (anonymous):

\[P(x=3)=10\times (.6)^3\times (.4)^2\] \[P(x=4)=5\times (.6)^4\times (.4)\] \[P(x=5)=(.6)^5\]

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!