An 8-sided fair die is rolled twice and the product of the two numbers obtained when the die is rolled two times is calculated.
1 answer:
<u>Question Completion</u>
(b) Use the possibility diagram to calculate the probability that the product of the two numbers is
I) A prime number
ii) Not a perfect square
iii) A multiple of 5
iv) Less than or equal to 21
v) Divisible by 4 or 6
Answer:
I) 0.125 (ii)0.828125 (iii)0.234375 (iv) 0.625 (v)0.65625
Step-by-step explanation:
The sample space for an 8-sided fair die rolled twice is:
(a) The possibility diagram of the product of the two numbers appearing on the die in each throw
(b)
I) A prime number
Number of Products which are prime numbers = 8
ii) Not a perfect square
Number of products which are perfect squares =11
iii) A multiple of 5
Number of Products that are multiples of 5=15
iv) Less than or equal to 21
Number of products less than or equal to 21=40
v) Divisible by 4 or 6
Products Divisible by 4 or 6= 42
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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,
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,
(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,
The probability that assembly is replaced the first day means the probability that at least one blade is dull is,
(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,
The geometric probability mass function is,
The probability that assembly is replaced no replaced until the third day of evaluation is,
(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.
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.
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.
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