Mathematics, 23.03.2020 17:04 naenae662

A company that produces fine crystal knows from experience that 13% of its goblets have cosmetic flaws and must be classified as "seconds." (Round your answers to four decimal places.)

(a)Among six randomly selected goblets, how likely is it that only one is a second?
(b)Among six randomly selected goblets, what is the probability that at least two are seconds?
(c)If goblets are examined one by one, what is the probability that at most five must be selected to find four that are not seconds?

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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:

$\rm P(X=x) = (^n_x)p^x(1-p)^{n-x}$

$\rm P(X=1) = (^6_1)(0.13)^1(1-0.13)^{6-1}$

Simplify the above expression.

P(X = 1) = 0.3888

b) The probability that at least two are seconds is given below:

$\rm P(x\geq 2) = 1-P(X$

$\rm P(X\geq 2) = 1- (^6_0)(0.13)^0(1-0.13)^{6-0}$

Simplify the above expression.

$\rm P(X\geq 2) = 0.1776$

c) The probability that at most five must be selected to find four that are not seconds is calculated as:

$\rm P(4 \;not\;seconds )= P(X=0;n=4)+P(X=1;n=5)$

$\rm = (^4_0)(0.13)^0(1-0.13)^{4-0}+ (^5_1)(0.13)^1(1-0.13)^{5-1}$

= 0.5729+0.3724

= 0.9453

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Answer from: aaroneduke4576

(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.

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:

$P(X=x)={n\choose x}p^{x}(1-p)^{n-x};\ x=0,1,2,3...$

(a)

Compute the probability that only one goblet is a second among six randomly selected goblets as follows:

$P(X=1)={6\choose 1}0.13^{1}(1-0.13)^{6-1}\\=6\times 0.13\times 0.4984\\=0.3888$

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)

$=1-{6\choose 0}0.13^{0}(1-0.13)^{6-0}-{6\choose 1}0.13^{1}(1-0.13)^{6-1}\\=1-0.4336+0.3888\\=0.1776$

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)

$={4\choose 0}0.13^{0}(1-0.13)^{4-0}+{5\choose 1}0.13^{1}(1-0.13)^{5-1}\\=0.5729+0.3724\\=0.9453$

Thus, the probability that at most five must be selected to find four that are not seconds is 0.9453.

Answer from: Quest

x is the total time in minutes for the free response

you are given number of questions as 15 - m and time per item as 8 minutes

to find x multiply the two givens together:

x = 8 * (15-m) = 8(15-m)

40 points students are given 3 minutes to complete each multiple-choice question on a test and 8 min

Answer from: Quest

im guessing (23,14). hope this : )

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A company that produces fine crystal knows from experience that 13% of its goblets have cosmetic fla...