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Answer:
Probability that 9 will have 0 defects is 0.0013
Probability that 4 will have 1 defect is 0.0274
Probability that 3 will have 2 defects is 0.0211
Step-by-step explanation:
We are given that before a television set leaves the factory, it is given a quality control check. The probability that a television contains 0, 1, or 2 defects is 0.88, 0.08, and 0.04, respectively.
A sample of 16 televisions is selected.
The above situation can be represented through Binomial distribution;
[tex]P(X=r) = \binom{n}{r}p^{r} (1-p)^{n-r} ; x = 0,1,2,3,.....[/tex]
where, n = number of trials (samples) taken = 16 televisions
r = number of success
p = probability of success which in our question is probability
of television containing defects
LET X = Number of televisions containing defects
(a) Now, in first part we have to find the probability that 9 will have 0 defects.
Here, p = probability that television contain 0 defects = 0.88
and r = 9
SO, X ~ Binom(n = 16, p = 0.88)
Probability that 9 will have 0 defects is given by = P(X = 9)
P(X = 9) = [tex]\binom{16}{9} \times 0.88^{9} \times (1-0.88)^{16-9}[/tex]
= [tex]11440 \times 0.88^{9}\times 0.12^{7}[/tex]
= 0.0013
(b) Now, in second part we have to find the probability that 4 will have 1 defect.
Here, p = probability that television contain 1 defect = 0.08
and r = 4
SO, X ~ Binom(n = 16, p = 0.08)
Probability that 4 will have 1 defect is given by = P(X = 4)
P(X = 4) = [tex]\binom{16}{4} \times 0.08^{4} \times (1-0.08)^{16-4}[/tex]
= [tex]1820 \times 0.08^{4}\times 0.92^{12}[/tex]
= 0.0274
(c) Now, in third part we have to find the probability that 3 will have 2 defects.
Here, p = probability that television contain 2 defects = 0.04
and r = 3
SO, X ~ Binom(n = 16, p = 0.04)
Probability that 3 will have 2 defects is given by = P(X = 3)
P(X = 3) = [tex]\binom{16}{3} \times 0.04^{3} \times (1-0.96)^{16-3}[/tex]
= [tex]560 \times 0.04^{3}\times 0.96^{13}[/tex]
= 0.0211
The required value is [tex]0.0003[/tex] (rounded).
To understand the calculations, check below.
Probability:
Probability is a measure of the likelihood of an event to occur. Many events cannot be predicted with total certainty. We can predict only the chance of an event to occur, i.e. how likely they are to happen, using it.
Clearly, it is multinomial distribution.
Here,
[tex]n=16\\X_{1}=9 \ \ \ P_{1}=0.88 \\X_{2}=4 \ \ \ P_2=0.08\\ X_3=3 \ \ \ P_3=0.04\\We observe that,\\X_1+X_2+X_3=16=n\\P_1+P_2+P_3=1\\[/tex]
Therefore, the required probability is given below,
[tex]\therefore P(X)=\frac{n!}{X_1!X_2!X_3!} P_1^X_1P_2^X_2P_3^X_3\\=\frac{16!}{9!4!3!} (0.88)^9(0.08)^4(0.04)^3\\=0.0003322\\=0.0003 \ (rounded)[/tex]
Learn more about the topic of Probability: https://brainly.com/question/6649771
