In a standard deck, there are 52 cards. Twelve cards are face cards (
F ) and 40 cards are not face cards (
N ). Draw two cards, one at a time, without replacement. The tree diagram is labeled with all possible probabilities.
Find
P (
FN OR
NF ).
Find
P (
N |
F ).
Find
P (at most one face card).
Hint: "At most one face card" means zero or one face card.
Find
P (at least on face card).
Hint: "At least one face card" means one or two face cards.
P (
FN OR
NF ) =
$\frac{\text{480}}{\text{2,652}}\text{+}\frac{\text{480}}{\text{2,652}}\text{=}\frac{\text{960}}{\text{2,652}}\text{=}\frac{\text{80}}{\text{221}}$
P (
N |
F ) =
$\frac{40}{51}$
P (at most one face card) =
$\frac{\text{(480+480+1,560)}}{\text{2,652}}$ =
$\frac{2,520}{2,652}$
P (at least one face card) =
$\frac{\text{(132+480+480)}}{\text{2,652}}$ =
$\frac{\text{1,092}}{\text{2,652}}$
A litter of kittens available for adoption at the Humane Society has four tabby kittens and five black kittens. A family comes in and randomly selects two kittens (without replacement) for adoption.
What is the probability that both kittens are tabby?
a.
$\left(\frac{1}{2}\right)\left(\frac{1}{2}\right)$ b.
$\left({\scriptscriptstyle \frac{4}{9}}\right)\left({\scriptscriptstyle \frac{4}{9}}\right)$ c.
$\left({\scriptscriptstyle \frac{4}{9}}\right)\left({\scriptscriptstyle \frac{3}{8}}\right)$ d.
$\left({\scriptscriptstyle \frac{4}{9}}\right)\left({\scriptscriptstyle \frac{5}{9}}\right)$
What is the probability that one kitten of each coloring is selected?
a.
$\left({\scriptscriptstyle \frac{4}{9}}\right)\left({\scriptscriptstyle \frac{5}{9}}\right)$ b.
$\left({\scriptscriptstyle \frac{4}{9}}\right)\left({\scriptscriptstyle \frac{5}{8}}\right)$ c.
$\left({\scriptscriptstyle \frac{4}{9}}\right)\left({\scriptscriptstyle \frac{5}{9}}\right)+\left({\scriptscriptstyle \frac{5}{9}}\right)\left({\scriptscriptstyle \frac{4}{9}}\right)$ d.
$\left({\scriptscriptstyle \frac{4}{9}}\right)\left({\scriptscriptstyle \frac{5}{8}}\right)+\left({\scriptscriptstyle \frac{5}{9}}\right)\left({\scriptscriptstyle \frac{4}{8}}\right)$
What is the probability that a tabby is chosen as the second kitten when a black kitten was chosen as the first?
What is the probability of choosing two kittens of the same color?
a. c, b. d, c.
$\frac{4}{8}$ , d.
$\frac{32}{72}$
Try it
Suppose there are four red balls and three yellow balls in a box. Three balls are drawn from the box without replacement. What is the probability that one ball of each coloring is selected?
A
Venn diagram is a picture that represents the outcomes of an experiment. It generally consists of a box that represents the sample space S together with circles or ovals. The circles or ovals represent events.
Suppose an experiment has the outcomes 1, 2, 3, ... , 12 where each outcome has an equal chance of occurring. Let event
A = {1, 2, 3, 4, 5, 6} and event
B = {6, 7, 8, 9}. Then
A AND
B = {6} and
A OR
B = {1, 2, 3, 4, 5, 6, 7, 8, 9}. The Venn diagram is as follows:
Try it
Suppose an experiment has outcomes black, white, red, orange, yellow, green, blue, and purple, where each outcome has an equal chance of occurring. Let event
C = {green, blue, purple} and event
P = {red, yellow, blue}. Then
C AND
P = {blue} and
C OR
P = {green, blue, purple, red, yellow}. Draw a Venn diagram representing this situation.
Flip two fair coins. Let
A = tails on the first coin. Let
B = tails on the second coin. Then
A = {
TT ,
TH } and
B = {
TT ,
HT }. Therefore,
A AND
B = {
TT }.
A OR
B = {
TH ,
TT ,
HT }.
The sample space when you flip two fair coins is
X = {
HH ,
HT ,
TH ,
TT }. The outcome
HH is in NEITHER
A NOR
B . The Venn diagram is as follows:
Try it
Roll a fair, six-sided die. Let
A = a prime number of dots is rolled. Let
B = an odd number of dots is rolled. Then
A = {2, 3, 5} and
B = {1, 3, 5}. Therefore,
A AND
B = {3, 5}.
A OR
B = {1, 2, 3, 5}. The sample space for rolling a fair die is
S = {1, 2, 3, 4, 5, 6}. Draw a Venn diagram representing this situation.
Forty percent of the students at a local college belong to a club and
50% work part time.
Five percent of the students work part time and belong to a club. Draw a Venn diagram showing the relationships. Let
C = student belongs to a club and
PT = student works part time.
If a student is selected at random, find
the probability that the student belongs to a club.
P (
C ) = 0.40
the probability that the student works part time.
P (
PT ) = 0.50
the probability that the student belongs to a club AND works part time.
P (
C AND
PT ) = 0.05
the probability that the student belongs to a club
given that the student works part time.
$P\text{(}C\text{|}PT\text{)}=\frac{P\text{(}C\text{AND}PT\text{)}}{P\text{(}PT\text{)}}=\frac{0.05}{0.50}=0.1$
the probability that the student belongs to a club
OR works part time.
P (
C OR
PT ) =
P (
C ) +
P (
PT ) -
P (
C AND
PT ) = 0.40 + 0.50 - 0.05 = 0.85
Questions & Answers
can someone help me with some logarithmic and exponential equations.
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
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.
Source:
OpenStax, Introduction to statistics i - stat 213 - university of calgary - ver2015revb. OpenStax CNX. Oct 21, 2015 Download for free at http://legacy.cnx.org/content/col11874/1.3
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