The probability that a certain machine turns out a defective item is 5%. Find the probabilities that in a set of 75 items: (a) Exactly 5 defective items (b) No defective items (c) At least one defective item (d) What is the expected value of the number of defective items? CANNOT BE EXCEL! Thank you!

Question

The probability that a certain machine turns out a defective item is 5%. Find the probabilities that in a set of 75 items:   (a) Exactly 5 defective items   (b) No defective items   (c) At least one defective item   (d) What is the expected value of the number of defective items?   CANNOT BE EXCEL! Thank you!

Durst37 · Accepted Answer

Step 1As per bartleby guidelines only first three subparts are to be solved. Please upload other parts separately.Given:Probability of defective item is, P(A)=0.05Number of items, n=75It is known that probability mass function of binomial distribution is,P(X=r)=nCr⋅pr⋅(1−p)n−rStep 2a) To find the probability of getting exactly 5 defective items.Let&#039;s take in this case, r=5P(X=r)=nCr⋅pr⋅(1−p)n−rP(X=5)=75C5⋅p5⋅(1−p)75−5P(X=5)=75!(75−5)!5!⋅(0.05)5⋅(1−0.05)75−5P(X=5)=75!(70)!5!⋅(0.05)5⋅(0.95)70P(X=5)=75×74×73×72×71×70!(70)!5×3×2×1⋅0.0000003125⋅(0.027583)P(X=5)=75×74×73×72×715×3×2⋅0.0000003125⋅(0.027583)P(X=5)=0.1488Hence, the probability of getting exactly 5 defective items is 0.1488Step 3b) To find the probability of getting no defective items.Let&#039;s take in this case, r=0.For r=0, the probability mass function reduces to,P(X=r)=(1−p)n−rP(X=0)=(1−0.05)75−0P(X=0)=(0.95)75P(X=0)=0.0213Hence, the probability of getting no defective items is 0.0213Step 4c) The probability of getting atleast one defective item is the complement of probability of no defective item.P(X=0)c=1−P(X=0)P(X=0)c=1−0.0213P(X=0)c=0.9787Hence the probability of getting atleast one defective item is 0.9787

Annie Levasseur · Accepted Answer

Step 1  The distribution of the number of defective items in this run is binomial with n=75 items and p=0.05 of each item being defective. The probability mass function of the binomial distribution is:  P(X=k)=(nCk)×(pk)×(1−p)n−k  where nCk is the number of combinations of k objects chosen from  n=n!k!×(n−k)!  1) In this case, we set k=5, n=75 and p=0.05, so:  P(X=5)=(75C5)×(0.055)×(0.95)70  P(X=5)=(75!5!×(70)!)×(0.055)×(0.95)70  P(X=5)=(75!5!×(70)!)×(0.055)×(0.95)70=0.1488  2) In this case, we set k = 0. The binomial probability mass function reduces to (1=p)n when k=0, so P(X=0)=(0.95)75=0.0213  3) The event that at least one item is defective is the compliment of the event that there are no defective items. The probability of a complimentary event happening is 1−P(original event), so the probability of at least one defective item is 1−0.0213=0.9787.

karton · Accepted Answer

Step 1 Let X denotes the number of defective items in a machine which follows binomial distribution with the probability of success 0.05 the number of items selected is 75. That is, X∼P(75, 0.05) The probability mass function of X is given below: P(X=x)=(nx)px(1−p)n−x a) Obtain the probability of getting exactly 5 defective items: P(X=5)=(755)0.055(1−0.05)75−5 =17,259,390×0.055(0.95)70 =0.1488 Thus, P(Exactly 5 defective items)=0.1488 Step 2 b) Obtain the probability of getting at least one defective item: P(X≥1)=1−P(X&amp;lt;1) =1−P(X=0) =1−(750)0.050(1−0.05)75−0 =1−(0.95)75 =1−0.0213 Thus, P(No defective items)=0.0213. c) Obtain the probability of getting at least one defective item: P(X≥1)=1−P(X&amp;lt;1) =1−P(X=0) =1−(750)0.050(1−0.05)75−0 =1−(0.95)75 =1−0.0213 =0.9787 Thus, P(at least one defective item)=0.9787

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