We now demonstrate the above results with a tree diagram.
Suppose a jar contains 3 red and 4 white marbles. If two marbles are drawn without replacement, find the following probabilities using a tree diagram.
The probability that both marbles are white.
The probability that the first marble is red and the second white.
The probability that one marble is red and the other white.
Let
$R$ be the event that the marble drawn is red, and let
$W$ be the event that the marble drawn is white.
We draw the following tree diagram.
Although the tree diagrams give us better insight into a problem, they are not practical for problems where more than two or three things are chosen. In such cases, we use the concept of combinations that we learned in
[link] . This method is best suited for problems where the order in which the objects are chosen is not important, and the objects are chosen without replacement.
Suppose a jar contains 3 red, 2 white, and 3 blue marbles. If three marbles are drawn without replacement, find the following probabilities.
$P\left(\text{Two red and one white}\right)$
$P\left(\text{One of each color}\right)$
$P\left(\text{None blue}\right)$
$P\left(\text{At least one blue}\right)$
Let us suppose the marbles are labeled as
${R}_{1}$ ,
${R}_{2}$ ,
${R}_{3}$ ,
${W}_{1}$ ,
${W}_{2}$ ,
${B}_{1}$ ,
${B}_{2}$ ,
${B}_{3}$ .
$P\left(\text{Two red and one white}\right)$
We analyze the problem in the following manner.
Since we are choosing 3 marbles from a total of 8, there are
$\mathrm{8C3}=\text{56}$ possible combinations. Of these 56 combinations, there are
$\mathrm{3C2}\times \mathrm{2C1}=6$ combinations consisting of 2 red and one white. Therefore,
$P\left(\text{Two red and one white}\right)=\frac{\mathrm{3C2}\times \mathrm{2C1}}{\mathrm{8C3}}=\frac{6}{\text{56}}$ .
$P\left(\text{One of each color}\right)$
Again, there are
$\mathrm{8C3}=\text{56}$ possible combinations. Of these 56 combinations, there are
$\mathrm{3C1}\times \mathrm{2C1}\times \mathrm{3C1}=\text{18}$ combinations consisting of one red, one white, and one blue. Therefore,
$P\left(\text{One of each color}\right)=\frac{\mathrm{3C1}\times \mathrm{2C1}\times \mathrm{3C1}}{\mathrm{8C3}}=\frac{\text{18}}{\text{56}}$ .
By "at least one blue marble," we mean the following: one blue marble and two non-blue marbles, or two blue marbles and one non-blue marble, or all three blue marbles. So we have to find the sum of the probabilities of all three cases.
$P\left(\text{At least one blue}\right)=P\left(\text{one blue, two non-blue}\right)+P\left(\text{two blue, one non-blue}\right)+P\left(\text{three blue}\right)$
$P\left(\text{At least one blue}\right)=\frac{\mathrm{3C1}\times \mathrm{5C2}}{\mathrm{8C3}}+\frac{\mathrm{3C2}\times \mathrm{5C1}}{\mathrm{8C3}}+\frac{\mathrm{3C3}}{\mathrm{8C3}}$
$P\left(\text{At least one blue}\right)=\text{30}/\text{56}+\text{15}/\text{56}+1/\text{56}=\text{46}/\text{56}=\text{23}/\text{28}$ .
Alternately,
we use the fact that
$P\left(E\right)=1-P\left({E}^{c}\right)$ .
If the event
$E=\text{At least one blue}$ , then
${E}^{c}=\text{None blue}$ .
But from part c of this example, we have
$\left({E}^{c}\right)=5/\text{28}$
Five cards are drawn from a deck. Find the probability of obtaining two pairs, that is, two cards of one value, two of another value, and one other card.
Let us first do an easier problem–the probability of obtaining a pair of kings and queens.
Since there are four kings, and four queens in the deck, the probability of obtaining two kings, two queens and one other card is
$P\left(\text{A pair of kings and queens}\right)=\frac{\mathrm{4C2}\times \mathrm{4C2}\times \text{44}\mathrm{C1}}{\text{52}\mathrm{C5}}$
To find the probability of obtaining two pairs, we have to consider all possible pairs.
Since there are altogether 13 values, that is, aces, deuces, and so on, there are
$\text{13}\mathrm{C2}$ different combinations of pairs.
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
Azam
anybody can imagine what will be happen after 100 years from now in nano tech world
Prasenjit
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
Azam
name doesn't matter , whatever it will be change... I'm taking about effect on circumstances of the microscopic world
Prasenjit
how hard could it be to apply nanotechnology against viral infections such HIV or Ebola?
Damian
silver nanoparticles could handle the job?
Damian
not now but maybe in future only AgNP maybe any other nanomaterials
Azam
Hello
Uday
I'm interested in Nanotube
Uday
this technology will not going on for the long time , so I'm thinking about femtotechnology 10^-15
Prasenjit
can nanotechnology change the direction of the face of the world
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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8. It is known that 80% of the people wear seat belts, and 5% of the people quit smoking last year. If 4% of the people who wear seat belts quit smoking, are the events, wearing a seat belt and quitting smoking, independent?
Mr. Shamir employs two part-time typists, Inna and Jim for his typing needs. Inna charges $10 an hour and can type 6 pages an hour, while Jim charges $12 an hour and can type 8 pages per hour. Each typist must be employed at least 8 hours per week to keep them on the payroll. If Mr. Shamir has at least 208 pages to be typed, how many hours per week should he employ each student to minimize his typing costs, and what will be the total cost?
At De Anza College, 20% of the students take Finite Mathematics, 30% take Statistics and 10% take both. What percentage of the students take Finite Mathematics or Statistics?