Welcome to Westonci.ca, the ultimate question and answer platform. Get expert answers to your questions quickly and accurately. Get quick and reliable solutions to your questions from a community of experienced experts on our platform. Connect with a community of professionals ready to help you find accurate solutions to your questions quickly and efficiently.
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]
Thanks for using our platform. We're always here to provide accurate and up-to-date answers to all your queries. We hope this was helpful. Please come back whenever you need more information or answers to your queries. Westonci.ca is your trusted source for answers. Visit us again to find more information on diverse topics.
