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Sagot :
Answer:
a) 0.2316 = 23.16% probability that 0 carry intestinal parasites.
b) 0.4005 = 40.05% probability that at least two individuals carry intestinal parasites.
Step-by-step explanation:
For each trout, there are only two possible outcomes. Either they carry intestinal parasites, or they do not. Trouts are independent. This means that we use the binomial probability distribution to solve this question.
Binomial probability distribution
The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
In which [tex]C_{n,x}[/tex] is the number of different combinations of x objects from a set of n elements, given by the following formula.
[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]
And p is the probability of X happening.
You know that in a specific population of rainbow trout 15% of the individuals carry intestinal parasites.
This means that [tex]p = 0.15[/tex]
Assume you obtain a random sample of 9 individuals from this population:
This means that [tex]n = 9[/tex]
a. Calculate the probability that __ (last digit of your ID number) carry intestinal parasites.
Last digit is 0, so:
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 0) = C_{9,0}.(0.15)^{0}.(0.85)^{9} = 0.2316[/tex]
0.2316 = 23.16% probability that 0 carry intestinal parasites.
b. Calculate the probability that at least two individuals carry intestinal parasites.
This is
[tex]P(X \geq 2) = 1 - P(X < 2)[/tex]
In which
[tex]P(X < 2) = P(X = 0) + P(X = 1)[/tex]
So
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 0) = C_{9,0}.(0.15)^{0}.(0.85)^{9} = 0.2316[/tex]
[tex]P(X = 1) = C_{9,1}.(0.15)^{1}.(0.85)^{8} = 0.3679[/tex]
[tex]P(X < 2) = P(X = 0) + P(X = 1) = 0.2316 + 0.3679 = 0.5995[/tex]
[tex]P(X \geq 2) = 1 - P(X < 2) = 1 - 0.5995 = 0.4005[/tex]
0.4005 = 40.05% probability that at least two individuals carry intestinal parasites.
We will see that the probabilities are:
- a) 3.8*10^-8
- b) 0.26
How to find the probability.
We know that 15% of the population have the parasites, then if we take a trout at random we have:
- a probability of 0.15 that it has parasites.
- a probability of 0.85 that it does not have parasites.
a) The probability that 9 of them carry intestinal parasites (so all of them have).
This is just the product of the individual probabilities, so we have 9 times a probability of 0.15, this gives:
P(9) = (0.15)^9 = 3.8*10^-8
b) Probability that 2 out of 9 have parasites.
Then we have 0.15 two times, and 0.85 seven times.
But we also need to take in account the permutations, the different groups of 2 trouts that we can make out of 9 trouts is given by:
[tex]C(9, 2) = \frac{9!}{(9-2)!*2!} = \frac{9*8}{2} = 36[/tex]
This means that there are 36 different pairs of 2 trouts that can be the ones with parasites.
Then the probability that 2 trouts have parasites is:
P(2) = 36*(0.15)^2*(0.85)^7 = 0.26
And just to be complete, the probabilty that x trouts have parasites is:
P(x) = C(9, x)*(0.15)^x*(0.85)^(9 - x)
If you want to learn more about probability, you can read:
https://brainly.com/question/251701
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