Respuesta :
Answer:
Part a
The probability that assembly is replaced the first day is 0.7069.
Part b
The probability that assembly is replaced no replaced until the third day of evaluation is 0.0607.
Part c
The probability that the assembly is not replaced until the third day of evaluation is 0.2811.
Step-by-step explanation:
Hypergeometric Distribution: A random variable x that represents number of success of the n trails without replacement and M represents number of success of the N trails without replacement is termed as the hypergeometric distribution. Moreover, it consists of fixed number of trails and also the two possible outcomes for each trail.
It occurs when there is finite population and samples are taken without replacement.
The probability distribution of the hyper geometric is,
[tex]P(x,N,n,M)=\frac{(\limits^M_x)(\imits^{N-M}_{n-x})}{(\limits^N_n)}[/tex]
Here x is the success in the sample of n trails, N represents the total population, n represents the random sample from the total population and M represents the success in the population.
Probability that at least one of the trail is succeed is,
[tex]P(x\geq1)=1-P(x<1)= 1-P(x=0)[/tex]
(a)
Compute the probability that the assembly is replaced the first day.
From the given information,
Let x be number of blades dull in the assembly are replaced.
Total number of blades in the assembly N = 48.
Number of blades selected at random from the assembly n= 5
Number of blades in an assembly dull is M = 10.
The probability mass function is,
[tex]P(X=x)=\frac{[\limits^M_x][\limits^{N-M}_{n-x}]}{[\limits^N_n]};x=0,1,2,...,n\\\\=\frac{[\limits^{10}_x][\limits^{48-10}_{5-x}]}{[\limits^{48}_5]}[/tex]
The probability that assembly is replaced the first day means the probability that at least one blade is dull is,
[tex]P(x\geq 1)=1- P(x<1)\\=1-P(x=0)\\=1-\frac{(\limits^10_0)(\limits^{48-10}_{5-0})}{\limits^{48}_5}\\\\=1-\frac{\limits^{38}_5}{\limits^{38}_5}\\\\= 1-0.2931= 0.7069[/tex]
(b)
From the given information,
Let x be number of blades dull in the assembly are replaced.
Total number of blades in the assembly N = 48
Number of blades selected at random from the assembly N = 5
Number of blades in an assembly dull is M = 10
From the information,
The probability that assembly is replaced (P) is 0.7069.
The probability that assembly is not replaced is (Q) is,
[tex]q=1-p\\= 1-0.7069= 0.2931[/tex]
The geometric probability mass function is,
[tex]P(X = x)= q^{x-1} p; x =1,2,....=(0.2931)^{x-1}(0.7069)
[/tex]
The probability that assembly is replaced no replaced until the third day of evaluation is,
[tex]P(X = 3)=(0.2931)^{3-1}(0.7069)\\=(0.2931)^2(0.7069)= 0.0607[/tex]
(c)
From the given information,
Let x be number of blades dull in the assembly are replaced.
Total number of blades in the assembly N = 48
Number of blades selected at random from the assembly n = 5
Suppose that on the first day of the evaluation two of the blades are dull then the probability that the assembly is not replaced is,
Here, number of blades in an assembly dull is M = 2.
[tex]P(x=0)=\frac{(\limits^2_0)(\limits^{48-2}_{5-0})}{\limits^{48}_5}\\\\=\frac{(\limits^{46}_5)}{(\limits^{48}_5)}\\\\= 0.8005[/tex]
Suppose that on the second day of the evaluation six of the blades are dull then the probability that the assembly is not replaced is,
Here, number of blades in an assembly dull is M = 6.
[tex]P(x=0)=\frac{(\limits^6_0)(\limits^{48-6}_{5-0})}{(\limits^{48}_5)}\\\\=\frac{(\limits^{42}_5}{(\limits^{48}_5)}\\\\= 0.4968[/tex]
Suppose that on the third day of the evaluation ten of the blades are dull then the probability that the assembly is not replaced is,
Here, number of blades in an assembly dull is M
= 10.
[tex]P(x\geq 1)=1- P(x<1)\\=1-P(x=0)\\=1-\frac{(\limits^10_0)(\limits^{48-10}_{5-0})}{\limits^{48}_5}\\\\=1-\frac{\limits^{38}_5}{\limits^{38}_5}\\\\= 1-0.2931= 0.7069[/tex]
The probability that the assembly is not replaced until the third day of evaluation is,
P(The assembly is not replaced until the third day)=P(The assembly is not replaced first day) x P(The assembly is not replaced second day) x P(The assembly is replaced third day)
=(0.8005)(0.4968)(0.7069)= 0.2811
Using the hypergeometric distribution, it is found that there is a:
a) 0.7069 = 70.69% probability that the assembly is replaced the first day it is evaluated.
b) 0.0607 = 6.07% probability that the assembly is not replaced until the third day of evaluation.
c) 0.2811 = 28.11% probability that the assembly is not replaced until the third day of evaluation.
The blades are chosen without replacement, which means that the hypergeometric distribution is used to solve this question.
Hypergeometric distribution:
[tex]P(X = x) = h(x,N,n,k) = \frac{C_{k,x}C_{N-k,n-x}}{C_{N,n}}[/tex]
[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]
The parameters are:
- x is the number of successes.
- N is the size of the population.
- n is the size of the sample.
- k is the total number of desired outcomes.
In this problem:
- There are 48 blades, hence [tex]N = 48[/tex]
- 5 are chosen, hence [tex]n = 5[/tex]
Item a:
10 of the items are dull, hence [tex]k = 10[/tex]
The probability is:
[tex]P(X \geq 1) = 1 - P(X = 0)[/tex]
In which:
[tex]P(X = x) = h(x,N,n,k) = \frac{C_{k,x}C_{N-k,n-x}}{C_{N,n}}[/tex]
[tex]P(X = 0) = h(0,48,5,10) = \frac{C_{10,0}C_{38,5}}{C_{48,5}} = 0.2931[/tex]
[tex]P(X \geq 1) = 1 - P(X = 0) = 1 - 0.2931 = 0.7069[/tex]
0.7069 = 70.69% probability that the assembly is replaced the first day it is evaluated.
Item b:
- Not replaced on the first two days, each with 0.2931 probability.
- Replaced on the third day, with 0.7069 probability.
Hence:
[tex]p = (0.2931)^2(0.7069) = 0.0607[/tex]
0.0607 = 6.07% probability that the assembly is not replaced until the third day of evaluation.
Item c:
- First day: P(X = 0) with k = 2.
- Second day: P(X = 0) with k = 6.
- Third day: [tex]P(X \geq 1)[/tex] with k = 10 -> 0.7069.
Hence:
[tex]p_1 = P(X = 0) = h(0,48,5,2) = \frac{C_{2,0}C_{46,5}}{C_{48,5}} = 0.8005[/tex]
[tex]p_2 = P(X = 0) = h(0,48,5,6) = \frac{C_{6,0}C_{42,5}}{C_{48,5}} = 0.4968[/tex]
[tex]p = 0.8005(0.4968)(0.7069) = 0.2811[/tex]
0.2811 = 28.11% probability that the assembly is not replaced until the third day of evaluation.
A similar problem is given at https://brainly.com/question/24826394
