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3 years ago

There are 4 ducks swimming in a pond. If 2 have green wings 2 have blue wings, what fraction of the ducks have green wings?

Mathematics

1 answer:

zhannawk [14.2K]3 years ago

5 0

2/4 which can then be simplified down to 1/2.

So 1/2 of the ducks have green wings.

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The number of failures of a testing instrument from contamination particles on the product is a Poisson random variable with a m

Mazyrski [523]

Answer:

The probability that the instrument does not fail in an 8-hour shift is $P(X=0) \approx 0.8659$

The probability of at least 1 failure in a 24-hour day is $P(X\geq 1 )\approx 0.3508$

Step-by-step explanation:

The probability distribution of a Poisson random variable X representing the number of successes occurring in a given time interval or a specified region of space is given by the formula:

$P(X)=\frac{e^{-\mu}\mu^x}{x!}$

Let X be the number of failures of a testing instrument.

We know that the mean $\mu = 0.018$ failures per hour.

(a) To find the probability that the instrument does not fail in an 8-hour shift, you need to:

For an 8-hour shift, the mean is $\mu=8\cdot 0.018=0.144$

$P(X=0)=\frac{e^{-0.144}0.144^0}{0!}\\\\P(X=0) \approx 0.8659$

(b) To find the probability of at least 1 failure in a 24-hour day, you need to:

For a 24-hour day, the mean is $\mu=24\cdot 0.018=0.432$

$P(X\geq 1 )=1-P(X=0)\\\\P(X\geq 1 )=1-\frac{e^{-0.432}0.432^0}{0!}\\\\P(X\geq 1 )\approx 0.3508$

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3 years ago

In a certain Algebra 2 class of 25 students, 13 of them play basketball and 11 of them play baseball. There are 4 students who p

Flauer [41]

Answer:

$\frac{4}{5}$

Step-by-step explanation:

Given: There are 2 classes of 25 students.

13 play basketball

11 play baseball.

4 play neither of sports.

Lets assume basketball as "a" and baseball as "b".

We know, probablity dependent formula; P(a∪b)= P(a)+P(b)-p(a∩b)

As given total number of student is 25

Now, subtituting the values in the formula.

⇒P(a∪b)= $\frac{13}{25} +\frac{11}{25} -\frac{4}{25}$

taking LCD as 25 to solve.

⇒P(a∪b)= $\frac{13\times 1+11\times 1-4\times 1}{25} = \frac{20}{25}$

∴ P(a∪b)= $\frac{4}{5}$

Hence, the probability that a student chosen randomly from the class plays both basketball and baseball is $\frac{4}{5}$ .

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