



Consider the class
$\{A,\phantom{\rule{0.166667em}{0ex}}B,\phantom{\rule{0.166667em}{0ex}}C,\phantom{\rule{0.166667em}{0ex}}D\}$ of events. Suppose
the probability that at least one of the events
A or
C occurs is
0.75 and the probability that at least one of the four events occurs is 0.90.Determine the probability that neither of the events
A or
C but
at least one of the events
B or
D occurs.
Use the pattern
$P(E\cup F)=P\left(E\right)+P\left({E}^{c}F\right)$ and
${(A\cup C)}^{c}={A}^{c}{C}^{c}$ .
$$P(A\cup C\cup B\cup D)=P(A\cup C)+P\left({A}^{c}{C}^{c}(B\cup D)\right),\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\text{so}\phantom{\rule{4.pt}{0ex}}\text{that}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}P\left({A}^{c}{C}^{c}(B\cup D)\right)=0.900.75=0.15$$
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 Use minterm maps to show which of the
following statements are true for any class
$\{A,\phantom{\rule{0.166667em}{0ex}}B,\phantom{\rule{0.166667em}{0ex}}C\}$ :

$$\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}A\cup {\left(BC\right)}^{c}=A\cup B\cup {B}^{c}{C}^{c}$$

$$\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}{(A\cup B)}^{c}={A}^{c}C\cup {B}^{c}C$$

$$\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}A\subset AB\cup AC\cup BC$$
 Repeat part (1) using indicator functions (evaluated on minterms).
 Repeat part (1) using the mprocedure minvec3 and MATLAB logical operations.
We use the MATLAB procedure, which displays the essential patterns.
minvec3
Variables are A, B, C, Ac, Bc, CcThey may be renamed, if desired.
E = A~(B&C);
F = AB(Bc&Cc);
disp([E;F])
1 1 1 0 1 1 1 1 % Not equal1 0 1 1 1 1 1 1
G = ~(AB);H = (Ac&C)(Bc&C);
disp([G;H])
1 1 0 0 0 0 0 0 % Not equal0 1 0 1 0 1 0 0
K = (A&B)(A&C)(B&C);
disp([A;K])
0 0 0 0 1 1 1 1 % A not contained in K0 0 0 1 0 1 1 1
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Use (1) minterm maps, (2) indicator functions (evaluated on minterms),
(3) the mprocedure minvec3 and MATLAB logical operations to show that

$$\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}A(B\cup {C}^{c})\cup {A}^{c}BC\subset A(BC\cup {C}^{c})\cup {A}^{c}B$$

$$\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}A\cup {A}^{c}BC=AB\cup BC\cup AC\cup A{B}^{c}{C}^{c}$$
We use the MATLAB procedure, which displays the essential patterns.
minvec3
Variables are A, B, C, Ac, Bc, CcThey may be renamed, if desired.
E = (A&(BCc))(Ac&B&C);
F = (A&((B&C)Cc))(Ac&B);
disp([E;F])
0 0 0 1 1 0 1 1 % E subset of F0 0 1 1 1 0 1 1
G = A(Ac&B&C);
H = (A&B)(B&C)(A&C)(A&Bc&Cc);
disp([G;H])
0 0 0 1 1 1 1 1 % G = H0 0 0 1 1 1 1 1
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Minterms for the events
$\{A,B,C,D\}$ , arranged as on a minterm map
are
0.0168 0.0072 0.0252 0.0108
0.0392 0.0168 0.0588 0.02520.0672 0.0288 0.1008 0.0432
0.1568 0.0672 0.2352 0.1008
What is the probability that three or more of the events occur on a trial? Of exactly two?
Of two or fewer?
We use mintable(4) and determine positions with correct number(s) of
ones (number of occurrences). An alternate is to use minvec4 and express theBoolean combinations which give the correct number(s) of ones.
npr02_04 Minterm probabilities are in pm. Use mintable(4)
a = mintable(4);s = sum(a); % Number of ones in each minterm position
P1 = (s>=3)*pm' % Select and add minterm probabilities
P1 = 0.4716P2 = (s==2)*pm'
P2 = 0.3728P3 = (s<=2)*pm'
P3 = 0.5284
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Minterms for the events
$\{A,B,C,D,E\}$ , arranged as on a minterm map
are
0.0216 0.0324 0.0216 0.0324 0.0144 0.0216 0.0144 0.0216
0.0144 0.0216 0.0144 0.0216 0.0096 0.0144 0.0096 0.01440.0504 0.0756 0.0504 0.0756 0.0336 0.0504 0.0336 0.0504
0.0336 0.0504 0.0336 0.0504 0.0224 0.0336 0.0224 0.0336
What is the probability that three or more of the events occur on a trial? Of exactly four?
Of three or fewer? Of either two or four?
We use mintable(5) and determine positions with correct number(s) of
ones (number of occurrences).
npr02_05 Minterm probabilities are in pm. Use mintable(5)
a = mintable(5);s = sum(a); % Number of ones in each minterm position
P1 = (s>=3)*pm' % Select and add minterm probabilities
P1 = 0.5380P2 = (s==4)*pm'
P2 = 0.1712P3 = (s<=3)*pm'
P3 = 0.7952P4 = ((s==2)(s==4))*pm'
P4 = 0.4784
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Questions & Answers
Application of nanotechnology in medicine
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please someone correct me if I'm wrong but I think one can use nanoparticles, specially silver nanoparticles for water treatment.
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yes that's correct
Professor
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Rafiq
industrial application...? mmm I think on the medical side as drug carrier, but you should go deeper on your research, I may be wrong
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LITNING
scanning tunneling microscope
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Anam
analytical skills graphene is prepared to kill any type viruses .
Anam
what is Nano technology ?
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Bob
The nanotechnology is as new science, to scale nanometric
brayan
nanotechnology is the study, desing, synthesis, manipulation and application of materials and functional systems through control of matter at nanoscale
Damian
Is there any normative that regulates the use of silver nanoparticles?
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Renato
What fields keep nano created devices from performing or assimulating ? Magnetic fields ? Are do they assimilate ?
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Adin
biomolecules are e building blocks of every organics and inorganic materials.
Joe
anyone know any internet site where one can find nanotechnology papers?
sciencedirect big data base
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Introduction about quantum dots in nanotechnology
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Source:
OpenStax, Applied probability. OpenStax CNX. Aug 31, 2009 Download for free at http://cnx.org/content/col10708/1.6
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