Respuesta :
Answer:
a) 0.88
b) 0.35
c) 0.0144
d) 0.2084
e) 0.7916
Step-by-step explanation:
a) The probability of a peanut being brown is 12/100 = 0.12. Hence the probability of it not being brown is 1-0.12 = 0.88
b) 12% of peanuts are brown, 23% are blue. So 35% are either blue or brown. The probability of a peanut being blue or brown is, therefore 35/100 = 0.35.
c) 12% of peanuts are red, so the probability of a peanut being red is 12/100 = 0.12. In order to calculate the probability of 2 peanuts being both red, we can assume that the proportion doesnt change dramatically after removing one peanut (because the number of peanuts is absurdly high. We can assume that we are replenishing the peanuts). To calculate the probability of 2 peanuts being both red, we need to power 0.12 by 2, hence the probability is 0.12² = 0.0144.
d) Again, we will assume that the probability doesnt change, because we replenish. The probability of a peanut being blue is 0.23. The probability of it not being blue is 0.77, so the probability of 6 peanuts not being blue is obtained from powering 0.77 by 6, hence it is 0.77⁶ = 0.2084
e) The event 'at least one peanut is blue' is te complementary event of 'none peanuts are blue', so the probability of this event is 1- 0.2084 = 0.7916
Using probability concepts, especially the binomial distribution, it is found that:
- a) 88% probability that a randomly selected M&M is not brown.
- b) 35% probability a randomly selected peanut M&M is blue or brown.
- c) 0.0144 = 1.44% probability that two randomly selected peanut M&M’s are both red.
- d) 0.2084 = 20.84% probability that none of them are blue.
- e) 0.7916 = 79.16% probability that at least one of them is blue.
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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 a success.
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Item a:
- 12% of the M&M's are brown.
- Thus, 100 - 12 = 88% probability that a randomly selected M&M is not brown.
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Item b:
- 12% are brown.
- 23% are blue.
- Then, 12% + 23% = 35% probability a randomly selected peanut M&M is blue or brown.
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Item c:
- Multiple M&M's, so binomial distribution.
- 2 are selected, thus [tex]n = 2[/tex].
- 12% are red, thus [tex]p = 0.12[/tex]
The probability that both are red P(X = 2), so:
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 2) = C_{2,2}.(0.12)^{2}.(0.88)^{0} = 0.0144[/tex]
0.0144 = 1.44% probability that two randomly selected peanut M&M’s are both red.
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Item d:
- 6 selected, thus [tex]n = 6[/tex]
- 23% are blue, thus [tex]p = 0.23[/tex]
The probability that none are blue is P(X = 0). So:
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 0) = C_{6,0}.(0.23)^{0}.(0.77)^{6} = 0.2084[/tex]
0.2084 = 20.84% probability that none of them are blue.
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Item e:
Using the previous item:
[tex]P(X \geq 1) = 1 - P(X = 0) = 1 - 0.2084 = 0.7916[/tex]
0.7916 = 79.16% probability that at least one of them is blue.
A similar problem is given at https://brainly.com/question/15557838
