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Featured 4 months ago

Finding the probability of *at least***or** **or** **or** all **Or** implies addition.)

*not* owning a dog is

The probability of

#2# people out of#5# having a dog is equivalent to the probability of#2# people having a dog and#3# peoplenothaving a dog.

#overbrace(3/5 * 3/5) ^ "dog" * overbrace(2/5 * 2/5 * 2/5) ^ "no dog"# However, we don't know which two people own dogs. It can be the first and second person, first and fifth person, fourth and fifth person, etc. Basically, it can be any set of

#2# people out of the#5# . Thus, we have to multiply the combination#color(white)(I)_5C_2# along with everything.A quick recap on combinations:

#color(white)(I)_5C_2 = ((5), (2)) = (5!)/(2! * (5-2)!) = 10# You can also plug

#color(white)(I)_5C_2# into your calculator, which is what I'll be doing below.So, the probability of

#2# people owning a dog is actually

#(3/5)^2 * (2/5)^3 * color(white)(I)_5C_2 = color(red)0.2304#

We can do the same thing for all the other possibilities.

Probability of

#3# people owning a dog:

#(3/5)^3 * (2/5)^2 * color(white)(I)_5C_3 = color(red)0.3456# Probability of

#4# people owning a dog:

#(3/5)^4 * (2/5)^1 * color(white)(I)_5C_4 = color(red)0.2592# Probability of

#5# people owning a dog:

#(3/5)^5 * (2/5)^0 * color(white)(I)_5C_5 = color(red)0.07776#

Now that we have found each of the individual probabilities, we can add them up:

#color(red)0.2304 + color(red)0.3456 + color(red)0.2592 + color(red)0.07776 = 0.91296#

Again, this is the probability that **or** **or** **or**

#0.91296 ~~ color(blue)(91.3%)#

Featured 2 months ago

Let

We can set up an equation to reflect the following: the **four** exams of

Eliminate the denominator by multiplying

The equation becomes,

Add the three scores,

Then subtract

Verify by substituting

Featured 1 month ago

See a solution process below:

If we divide

If we divide the 36 shawls by 12 days we can determine how many shawls 1 person can weave in 1 day:

Now we can multiply this by 15 to find how many shawls 1 person can weave in 15 days:

Now we can multiply this by

The answers in **(C)** 6 people can weave 270 shawls in 15 days.

Featured 3 weeks ago

A)

B)

Questions such as these which provide an "average" or mean number of events in a given duration of time are typically handled using distributions known as Poisson Distributions. There are some conditions we must check before using a Poisson Distribution:

- We must be given (or know) the mean number of events happening in a specific unit of time.
- We must be able to count the observed number of events in the same unit of time if we were to watch for them, and these counts must be from the natural number set
#NN# . - The likelihood of events happening within one specific unit of time must be independent of the likelihood of those kind of events happening in a different unit of time of the same duration.

In this problem, we are talking about accidents. We know that the mean number of accidents happens at a rate of .4 accidents/day. For our problem, we use a variable

Let us use the variable

According to the Poisson formulas, the probability of

For the first question, we want to know what the probability is that we will go more than 3 days without an accident. Since our original time duration was a single day (with .4 mean accidents happening during that time), we could adjust that to represent a Poisson distribution that has a duration of 3 days. This leads us to a new distribution

Now we determine what the probability is of having 0 accidents in that Poisson distribution:

For the 2nd question we can proceed in a similar fashion. We're asked about the probability of going more than 5 days to get to the 3rd accident. We will derive a Poisson distribution of duration 5 days, or

In this distribution,

Featured 2 weeks ago

**With replacement**:

We can use the **negative binomial probability distribution.**

- For this type of probability distribution, an experiment is repeated several times. Let
#Y# be the number of the trial on which the#r^"th"# success occurs. Then#Y# is a**negative binomial random variable**, with parameters: probability#p# is the probability of success on only one trial and#r# is the number of successes.

Here the "experiment" is selecting bulbs, and a "success" is choosing a defective bulb. Then

The probability of selecting a defective bulb is

The probability distribution formula is given by:

We then have:

#=(1)(0.25)^2(1)(0.25)#

#=(0.25)^3#

#=0.015625#

A considerably low probability, which makes sense; we would be surprised if we managed to pick all three defective bulbs in the first three random attempts.

**Without replacement**, we use the **hypergeometric distribution.**

- For this type of probability distribution, a
**population**consists of N items, where each item is one of two types: there are#r# items of type one and the remaining#N-r# items are of type two. We select at random#n# of the#N# items, without replacement. Let#Y# be the number of items in the sample of#n# items that are type one. Then#Y# is a**hypergeometric random variable**, with parameters#N,r,n# .

Here the population is the total number of lights, so we have

The hypergeometric probability distribution formula is given as:

Then we have:

#P(Y=3)=((1)(1))/220#

#=0.00bar45#

Featured 1 week ago

Only the following two combinations are possible:

I will assume that both the median and the mean are

The median is the value in the middle of a set of data arranged in order. There are

The mode is the value that occurs the most often. The mode is

If the mean is

We already have a total of

Any combination of numbers bigger than 4 will do, except they cannot both be equal to

Only the following two combinations are possible:

The following are not possible:

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