Answer :
Answer:
(a) The probability that only one goblet is a second among six randomly selected goblets is 0.3888.
(b) The probability that at least two goblet is a second among six randomly selected goblets is 0.1776.
(c) The probability that at most five must be selected to find four that are not seconds is 0.9453.
Step-by-step explanation:
Let X = number of seconds in the batch.
The probability of the random variable X is, p = 0.31.
The random variable X follows a Binomial distribution with parameters n and p.
The probability mass function of X is:
[tex]P(X=x)={n\choose x}p^{x}(1-p)^{n-x};\ x=0,1,2,3...[/tex]
(a)
Compute the probability that only one goblet is a second among six randomly selected goblets as follows:
[tex]P(X=1)={6\choose 1}0.13^{1}(1-0.13)^{6-1}\\=6\times 0.13\times 0.4984\\=0.3888[/tex]
Thus, the probability that only one goblet is a second among six randomly selected goblets is 0.3888.
(b)
Compute the probability that at least two goblet is a second among six randomly selected goblets as follows:
P (X ≥ 2) = 1 - P (X < 2)
[tex]=1-{6\choose 0}0.13^{0}(1-0.13)^{6-0}-{6\choose 1}0.13^{1}(1-0.13)^{6-1}\\=1-0.4336+0.3888\\=0.1776[/tex]
Thus, the probability that at least two goblet is a second among six randomly selected goblets is 0.1776.
(c)
If goblets are examined one by one then to find four that are not seconds we need to select either 4 goblets that are not seconds or 5 goblets including only 1 second.
P (4 not seconds) = P (X = 0; n = 4) + P (X = 1; n = 5)
[tex]={4\choose 0}0.13^{0}(1-0.13)^{4-0}+{5\choose 1}0.13^{1}(1-0.13)^{5-1}\\=0.5729+0.3724\\=0.9453[/tex]
Thus, the probability that at most five must be selected to find four that are not seconds is 0.9453.
The probability that among six randomly selected goblets only one is a second is 0.3888, the probability that at least two are seconds is 0.1776, and the probability that at most five must be selected to find four that are not seconds is 0.9453.
Given :
A company that produces fine crystal knows from experience that 13% of its goblets have cosmetic flaws and must be classified as "seconds".
a) The probability that among six randomly selected goblets only one is a second is given below:
[tex]\rm P(X=x) = (^n_x)p^x(1-p)^{n-x}[/tex]
[tex]\rm P(X=1) = (^6_1)(0.13)^1(1-0.13)^{6-1}[/tex]
Simplify the above expression.
P(X = 1) = 0.3888
b) The probability that at least two are seconds is given below:
[tex]\rm P(x\geq 2) = 1-P(X<2)[/tex]
[tex]\rm P(X\geq 2) = 1- (^6_0)(0.13)^0(1-0.13)^{6-0}[/tex]
Simplify the above expression.
[tex]\rm P(X\geq 2) = 0.1776[/tex]
c) The probability that at most five must be selected to find four that are not seconds is calculated as:
[tex]\rm P(4 \;not\;seconds )= P(X=0;n=4)+P(X=1;n=5)[/tex]
[tex]\rm = (^4_0)(0.13)^0(1-0.13)^{4-0}+ (^5_1)(0.13)^1(1-0.13)^{5-1}[/tex]
[tex]= 0.5729+0.3724[/tex]
[tex]= 0.9453[/tex]
For more information, refer to the link given below:
https://brainly.com/question/795909