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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
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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]
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