The probability that a certain machine turns out a defective item is 5%. Find the...

Maria Huey

Maria Huey

Answered

2021-12-27

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!

Answer & Explanation

Durst37

Durst37

Expert

2021-12-28Added 37 answers

Step 1
As per bartleby guidelines only first three subparts are to be solved. Please upload other parts separately.
Given:
Probability of defective item is, P(A)=0.05
Number of items, n=75
It is known that probability mass function of binomial distribution is,
P(X=r)=nCrpr(1p)nr
Step 2
a) To find the probability of getting exactly 5 defective items.
Let's take in this case, r=5
P(X=r)=nCrpr(1p)nr
P(X=5)=75C5p5(1p)755
P(X=5)=75!(755)!5!(0.05)5(10.05)755
P(X=5)=75!(70)!5!(0.05)5(0.95)70
P(X=5)=75×74×73×72×71×70!(70)!5×3×2×10.0000003125(0.027583)
P(X=5)=75×74×73×72×715×3×20.0000003125(0.027583)
P(X=5)=0.1488
Hence, the probability of getting exactly 5 defective items is 0.1488
Step 3
b) To find the probability of getting no defective items.
Let's take in this case, r=0.
For r=0, the probability mass function reduces to,
P(X=r)=(1p)nrP(X=0)=(10.05)750
P(X=0)=(0.95)75
P(X=0)=0.0213
Hence, the probability of getting no defective items is 0.0213
Step 4
c) The probability of getting atleast one defective item is the complement of probability of no defective item.
P(X=0)c=1P(X=0)
P(X=0)c=10.0213
P(X=0)c=0.9787
Hence the probability of getting atleast one defective item is 0.9787
Annie Levasseur

Annie Levasseur

Expert

2021-12-29Added 30 answers

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)×(1p)nk
where nCk is the number of combinations of k objects chosen from
n=n!k!×(nk)!
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 1P(original event), so the probability of at least one defective item is 10.0213=0.9787.
karton

karton

Expert

2022-01-04Added 439 answers

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, XP(75, 0.05)
The probability mass function of X is given below:
P(X=x)=(nx)px(1p)nx
a) Obtain the probability of getting exactly 5 defective items:
P(X=5)=(755)0.055(10.05)755
=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(X1)=1P(X<1)
=1P(X=0)
=1(750)0.050(10.05)750
=1(0.95)75
=10.0213
Thus, P(No defective items)=0.0213.
c) Obtain the probability of getting at least one defective item:
P(X1)=1P(X<1)
=1P(X=0)
=1(750)0.050(10.05)750
=1(0.95)75
=10.0213
=0.9787
Thus, P(at least one defective item)=0.9787

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