At Westonci.ca, we provide reliable answers to your questions from a community of experts. Start exploring today! Our platform provides a seamless experience for finding reliable answers from a network of experienced professionals. Connect with a community of professionals ready to provide precise solutions to your questions quickly and accurately.
Sagot :
Using the hypergeometric distribution, the probabilities are given as follows:
a) 0.0333 = 3.33%.
b) 0.5 = 50%.
c) 0.8333 = 83.33%.
d) 0.9667 = 96.67%.
What is the hypergeometric distribution formula?
The formula is:
[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.
The values of the parameters are:
N = 10, n = 3, k = 4.
Item a:
The probability is P(X = 3), hence:
[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 = 3) = h(3,10,3,4) = \frac{C_{4,3}C_{6,0}}{C_{10,3}} = 0.0333[/tex]
Item b:
The probability is P(X = 1), hence:
[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 = 1) = h(1,10,3,4) = \frac{C_{4,1}C_{6,2}}{C_{10,3}} = 0.5[/tex]
Item c:
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(0,10,3,4) = \frac{C_{4,0}C_{6,3}}{C_{10,3}} = 0.1667[/tex]
Then:
[tex]P(X \geq 1) = 1 - P(X = 0) = 1 - 0.1667 = 0.8333[/tex]
Item d:
The probability is:
[tex]P(X \leq 2) = 1 - P(X = 3)[/tex].
Considering item a:
[tex]P(X \leq 2) = 1 - P(X = 3) = 1 - 0.0333 = 0.9667[/tex]
More can be learned about the hypergeometric distribution at https://brainly.com/question/24826394
#SPJ1
Let [tex]O[/tex] be the random variable for the number of bad oranges picked out. Then
[tex]P(O=o) = \dfrac{\dbinom6{3-o} \dbinom4o}{\dbinom{10}3}[/tex]
if [tex]o\in\{0,1,2,3\}[/tex] and zero otherwise, where
[tex]\dbinom nk = \dfrac{n!}{k!(n-k)!}[/tex]
is the so-called binomial coefficient.
(a) Of the 6 good oranges, you pick 0. Of the 3 bad oranges, you pick 3. Of the 10 total oranges, you pick 3. So the probability of picking out all bad oranges is
[tex]P(O=3) = \dfrac{\dbinom60 \dbinom43}{\dbinom{10}3} = \boxed{\dfrac1{30}}[/tex]
(b) By similar reasoning, the probability of drawing exactly 1 bad orange is
[tex]P(O=1) = \dfrac{\dbinom62 \dbinom41}{\dbinom{10}3} = \boxed{\dfrac12}[/tex]
(c) "At least 1 bad orange" means you pick out 1, 2, or 3 bad oranges. These events are mutually exclusive, and we already know the probabilities of picking out exactly 1 or all 3 bad oranges. The remaining probability of drawing 2 bad oranges is
[tex]P(O=2) = \dfrac{\dbinom61 \dbinom42}{\dbinom{10}3} = \dfrac3{10}[/tex]
so the overall probability of drawing at least 1 bad orange is
[tex]P(O\ge1) = P(O=1)+P(O=2)+P(O=3) =\dfrac1{30} + \dfrac3{10} + \dfrac12 = \boxed{\dfrac56}[/tex]
(d) I assume you mean "at most 2 bad oranges," meaning you pick out 0, 1, or 2 bad oranges. Again, these events are mutually exclusive, and the probability of picking out no bad oranges is
[tex]P(O=0) = \dfrac{\dbinom63 \dbinom40}{\dbinom{10}3} = \dfrac16[/tex]
hence the probability of drawings at most 2 bad oranges is
[tex]P(O\le2) = P(O=0) + P(O=1) + P(O=2) = \dfrac16 + \dfrac12 + \dfrac3{10} = \boxed{\dfrac{29}{30}}[/tex]
Thank you for visiting our platform. We hope you found the answers you were looking for. Come back anytime you need more information. Thank you for your visit. We're committed to providing you with the best information available. Return anytime for more. We're here to help at Westonci.ca. Keep visiting for the best answers to your questions.
