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Given events J and K: P(J) = 0.18 ; P(K) = 0.37 ; P(J or K) = 0.45

  • Find P(J and K)
  • Find the probability of the complement of event (J and K)
  • Find the probability of the complement of event (J or K)
  • P(J or K) = P(J) + P(K) − P(J and K); 0.45 = 0.18 + 0.37 − P(J and K) ; solve to find P(J and K) = 0.10
  • P( NOT (J and K) ) = 1 − P(J and K) = 1 − 0.10 = 0.90
  • P( NOT (J or K) ) = 1 − P(J or K) = 1 − 0.45 = 0.55
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United Blood Services is a blood bank that serves more than 500 hospitals in 18 states. According to their website, http://www.unitedbloodservices.org/humanbloodtypes.html, a person with type O blood and a negative Rh factor (Rh−) can donate blood to any person with any bloodtype. Their data show that 43% of people have type O blood and 15% of people have Rh− factor; 52% of people have type O or Rh− factor.

  • Find the probability that a person has both type O blood and the Rh− factor
  • Find the probability that a person does NOT have both type O blood and the Rh− factor.
  • P(Type O or Rh−) = P(Type O) + P(Rh−) − P(Type O and Rh−)
    0.52 = 0.43 + 0.15 − P(Type O and Rh−); solve to find P(Type O and Rh−) = 0.06
    6% of people have type O Rh− blood
  • P( NOT (Type O and Rh−) ) = 1 − P(Type O and Rh−) = 1 − 0.06 = 0.94
    94% of people do not have type O Rh− blood
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At a college, 72% of courses have final exams and 46% of courses require research papers. Suppose that 32% of courses have a research paper and a final exam. Let F be the event that a course has a final exam. Let R be the event that a course requires a research paper.

  • Find the probability that a course has a final exam or a research project.
  • Find the probability that a course has NEITHER of these two requirements.
  • P(R or F) = P(R) + P(F) − P(R and F) = 0.72 + 0.46 − 0.32 = 0.86
  • P( Neither R nor F ) = 1 − P(R or F) = 1 − 0.86 = 0.14
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In a box of assorted cookies, 36% contain chocolate and 12% contain nuts. Of those, 8% contain both chocolate and nuts. Sean is allergic to both chocolate and nuts.

  • Find the probability that a cookie contains chocolate or nuts (he can't eat it).
  • Find the probability that a cookie does not contain chocolate or nuts (he can eat it).
  • Let C be the event that the cookie contains chocolate. Let N be the event that the cookie contains nuts.
  • P(C or N) = P(C) + P(N) − P(C and N) = 0.36 + 0.12 − 0.08 = 0.40
  • P( neither chocolate nor nuts) = 1 − P(C or N) = 1 − 0.40 = 0.60
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A college finds that 10% of students have taken a distance learning class and that 40% of students are part time students. Of the part time students, 20% have taken a distance learning class. Let D = event that a student takes a distance learning class and E = event that a student is a part time student

  • Find P(D and E)
  • Find P(E | D)
  • Find P(D or E)
  • Using an appropriate test, show whether D and E are independent.
  • Using an appropriate test, show whether D and E are mutually exclusive.
  • P(D and E) = P(D|E)P(E) = (0.20)(0.40) = 0.08
  • P(E|D) = P(D and E) / P(D) = 0.08/0.10 = 0.80
  • P(D or E) = P(D) + P(E) − P(D and E) = 0.10 + 0.40 − 0.08 = 0.42
  • Not Independent: P(D|E) = 0.20 which does not equal P(D) = .10
  • Not Mutually Exclusive: P(D and E) = 0.08 ; if they were mutually exclusive then we would need to have P(D and E) = 0, which is not true here.
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When the Euro coin was introduced in 2002, two math professors had their statistics students test whether the Belgian 1 Euro coin was a fair coin. They spun the coin rather than tossing it, and it was found that out of 250 spins, 140 showed a head (event H) while 110 showed a tail (event T). Therefore, they claim that this is not a fair coin.

  • Based on the data above, find P(H) and P(T).
  • Use a tree to find the probabilities of each possible outcome for the experiment of tossing the coin twice.
  • Use the tree to find the probability of obtaining exactly one head in two tosses of the coin.
  • Use the tree to find the probability of obtaining at least one head.

  • P(H) = 140/250; P(T) = 110/250
  • 308/625
  • 504/625

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A box of cookies contains 3 chocolate and 7 butter cookies. Miguel randomly selects a cookie and eats it. Then he randomly selects another cookie and eats it also. (How many cookies did he take?)

  • Draw the tree that represents the possibilities for the cookie selections. Write the probabilities along each branch of the tree.
  • Are the probabilities for the flavor of the SECOND cookie that Miguel selects independent of his first selection? Explain.
  • For each complete path through the tree, write the event it represents and find the probabilities.
  • Let S be the event that both cookies selected were the same flavor. Find P(S).
  • Let T be the event that both cookies selected were different flavors. Find P(T) by two different methods: by using the complement rule and by using the branches of the tree. Your answers should be the same with both methods.
  • Let U be the event that the second cookie selected is a butter cookie. Find P(U).
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**Exercises 33 - 40 contributed by Roberta Bloom

Questions & Answers

can someone help me with some logarithmic and exponential equations.
Jeffrey Reply
sure. what is your question?
okay, so you have 6 raised to the power of 2. what is that part of your answer
I don't understand what the A with approx sign and the boxed x mean
it think it's written 20/(X-6)^2 so it's 20 divided by X-6 squared
I'm not sure why it wrote it the other way
I got X =-6
ok. so take the square root of both sides, now you have plus or minus the square root of 20= x-6
oops. ignore that.
so you not have an equal sign anywhere in the original equation?
Commplementary angles
Idrissa Reply
im all ears I need to learn
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what is a good calculator for all algebra; would a Casio fx 260 work with all algebra equations? please name the cheapest, thanks.
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a perfect square v²+2v+_
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algebra 2 Inequalities:If equation 2 = 0 it is an open set?
Kim Reply
or infinite solutions?
The answer is neither. The function, 2 = 0 cannot exist. Hence, the function is undefined.
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if |A| not equal to 0 and order of A is n prove that adj (adj A = |A|
Nancy Reply
rolling four fair dice and getting an even number an all four dice
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Kristine 2*2*2=8
Bridget Reply
Differences Between Laspeyres and Paasche Indices
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No. 7x -4y is simplified from 4x + (3y + 3x) -7y
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is it 3×y ?
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J, combine like terms 7x-4y
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how do you translate this in Algebraic Expressions
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Need to simplify the expresin. 3/7 (x+y)-1/7 (x-1)=
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. After 3 months on a diet, Lisa had lost 12% of her original weight. She lost 21 pounds. What was Lisa's original weight?
Chris Reply
what's the easiest and fastest way to the synthesize AgNP?
Damian Reply
types of nano material
abeetha Reply
I start with an easy one. carbon nanotubes woven into a long filament like a string
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what is nanomaterials​ and their applications of sensors.
Ramkumar Reply
what is nano technology
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what is system testing?
preparation of nanomaterial
Victor Reply
Yes, Nanotechnology has a very fast field of applications and their is always something new to do with it...
Himanshu Reply
good afternoon madam
what is system testing
what is the application of nanotechnology?
In this morden time nanotechnology used in many field . 1-Electronics-manufacturad IC ,RAM,MRAM,solar panel etc 2-Helth and Medical-Nanomedicine,Drug Dilivery for cancer treatment etc 3- Atomobile -MEMS, Coating on car etc. and may other field for details you can check at Google
anybody can imagine what will be happen after 100 years from now in nano tech world
after 100 year this will be not nanotechnology maybe this technology name will be change . maybe aftet 100 year . we work on electron lable practically about its properties and behaviour by the different instruments
name doesn't matter , whatever it will be change... I'm taking about effect on circumstances of the microscopic world
how hard could it be to apply nanotechnology against viral infections such HIV or Ebola?
silver nanoparticles could handle the job?
not now but maybe in future only AgNP maybe any other nanomaterials
can nanotechnology change the direction of the face of the world
Prasenjit Reply
At high concentrations (>0.01 M), the relation between absorptivity coefficient and absorbance is no longer linear. This is due to the electrostatic interactions between the quantum dots in close proximity. If the concentration of the solution is high, another effect that is seen is the scattering of light from the large number of quantum dots. This assumption only works at low concentrations of the analyte. Presence of stray light.
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the Beer law works very well for dilute solutions but fails for very high concentrations. why?
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1 It is estimated that 30% of all drivers have some kind of medical aid in South Africa. What is the probability that in a sample of 10 drivers: 3.1.1 Exactly 4 will have a medical aid. (8) 3.1.2 At least 2 will have a medical aid. (8) 3.1.3 More than 9 will have a medical aid.
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Source:  OpenStax, Collaborative statistics. OpenStax CNX. Jul 03, 2012 Download for free at http://cnx.org/content/col10522/1.40
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