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Answer:

#91.3%#

Explanation:

Finding the probability of at least #2# of the #5# people having a dog means that you have to find the probability of #2# people having a dog or #3# people having a dog or #4# people having a dog or all #5# people having a dog. To do this, find each of the individual probabilities and find their sum. (Or implies addition.)

#60%# of people own a dog, so the probability that any one person has a dog is also #60%# or #3/5#. The probability of someone not owning a dog is #1 - 3/5 = 2/5#.

The probability of #2# people out of #5# having a dog is equivalent to the probability of #2# people having a dog and #3# people not having 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 #2# people have a dog or #3# people have a dog or #4# people have a dog or #5# people have a dog, which is the same thing as finding the probability of at least #2# people having a dog.

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

Answer:

#84# is the score of the fourth test.

Explanation:

Let #x# represent the score for the fourth test since it is unknown.

We can set up an equation to reflect the following: the #3# test scores, the fourth score #x# and the average of all four exams of #85#.

#(86 + 78 + 92 + x)/4 = 85#

Eliminate the denominator by multiplying #4# on both sides of the equation:

#4 xx((86 + 78 + 92 + x)/4) = 85 xx 4#

#color(red)cancel(4) xx((86 + 78 + 92 + x)/color(red)cancel(4)) = 85 xx 4#

The equation becomes,

#86 + 78 + 92 + x = 340#

Add the three scores,

#256 + x = 340#

Then subtract #256# from both sides to determine #x#,

#256-256 + x = 340-256#

#x = 84# is the score of the fourth test.

Verify by substituting #x = 84# back into the original equation:

#(86 + 78 + 92 + x)/4 = 85" "color(red)rarr" "(86 + 78 + 92 + 84)/4 = 85#

#340/4 = 85#

#85 = 85# is balanced and therefore #x = 84# is valid.

Answer:

See a solution process below:

Explanation:

If we divide #180# by #5# we can determine how many shawls #1# person can weave in 12 days:

#180/5 = 36# Therefore, 1 person can weave 36 shawls in 12 days.

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

#36/12 = 3# Therefore, 1 person can weave 3 shawls per day.

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

#3 xx 15 = 45# 1 person can weave 45 shawls in 15 days.

Now we can multiply this by #6# to find out how many shawls 6 people can weave in 15 days:

#45 xx 6 = 270#

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

Answer:

A) #P[X = 0 | X~"Poi"(lambda=1.2)] = 0.301 or 30.1%#
B) #P[X < 3 | X~"Poi"(lambda=2)] = 0.6767 or 67.67%#

Explanation:

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 #lambda# (lambda) to store this value. Thus, #lambda = .4#.

Let us use the variable #X# to represent the number of accidents that will happen in one day. We will say that the probabilities of #X# happening are distributed according to a Poisson Distribution of parameter #lambda = .4#, or more commonly #X~"Poi"(lambda=.4)#.

According to the Poisson formulas, the probability of #x# events happening in the same length of time is given by the formula:

#P(X = x)=(lambda^x)/(x!e^lambda)#

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 #X~"Poi"(lambda=1.2)#, since .4 accidents/day leads to an average of (.4)(3) = 1.2 accidents/3 days.

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

#P(X = 0) = lambda^0/(0!e^lambda) = 1.2^0/(e^1.2)=1/e^1.2~~0.301#

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 #X~"Poi"(lambda=2)#. (This is because (.4)(5) = 2).

In this distribution, #X# represents the number of accidents in 5 days, and since we'd like that to be less than 3, we seek #P(X<3)=sum_(x=0)^2P(X = x)#:

#=sum_(x=0)^2 lambda^x/(x!e^lambda) =2^0/(0!e^2)+2^1/(1!e^2)+2^2/(2!e^2)#
#color(white)(=sum_(x=0)^2 lambda^x/(x!e^lambda))~~0.1353+0.2707+0.2707#
#color(white)(=sum_(x=0)^2 lambda^x/(x!e^lambda))=0.6767#

Answer:

#P(Y=3)=1/64# with replacement or #1/220# without replacement.

Explanation:

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 #Y=3# is the number of the trial on which the third success occurs, where #r=3#.

The probability of selecting a defective bulb is #3/12=1/4=0.25#. Therefore #q#, the probability of "failure" is #1-p=0.75#.

The probability distribution formula is given by:

enter image source here

We then have:

enter image source here

enter image source here

#=(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 #N=12#. Type one is a defective bulb, so #r=3#, and hence the number of non-defective bulbs is given by #12-3=9#. Additionally, we have #n=3#.

The hypergeometric probability distribution formula is given as:

enter image source here

Then we have:

enter image source here

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

#=0.00bar45#

Answer:

Only the following two combinations are possible:

#1" " , 1" ", 4" ", 5" "9#

#1" " , 1" ", 4" ", 6" "8#

Explanation:

I will assume that both the median and the mean are #4#.

The median is the value in the middle of a set of data arranged in order. There are #5# numbers, so the third one has to be #4#

#--- , ---, 4 ---, ---" "larr# the median is #4#

The mode is the value that occurs the most often. The mode is #1#. There are only two possible spaces for #1#, so there must be two of them.

#1" " , 1" ", 4 ---, ---" "larr# the mode is #1#

If the mean is #4# and there are five numbers, it means that the total of the five numbers is #5 xx 4 =20#

We already have a total of #6#, so the sum of the remaining teo numbers must be #20-6 =14#

Any combination of numbers bigger than 4 will do, except they cannot both be equal to #7#, because then there would be two modes.

Only the following two combinations are possible:

#1" " , 1" ", 4" ", 5" "9#

#1" " , 1" ", 4" ", 6" "8#

The following are not possible:

#1" " , 1" ", 4" ", 4" "10" "larr# there are two modes

#1" " , 1" ", 4" ",7" "7" "larr# there are two modes

#1" " , 1" ", 3" ", 4" "11" "larr# the median is not #4#

#1" " , 1" ", 2" ", 4" "12" "larr# the median is not #4#

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