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?
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:
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:
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
But can't be a binomial because, the x numbers are 0 to 6, instead those would be "0" or "1" in a straight way
Nelson
You can do a chi-square test, but the assumption has to be a normal distribution, and the last f's number need to be "64"
Nelson
sorry the last f's numbers : "6 and 4" which are the observed values for 5 and 6 (expected values)
Nelson
hi
rajendra
can't understand basic of statistics ..
rajendra
Sorry I see my mistake, we have to calculate the expected values
Nelson
So we need this equation:
P= (X=x)=(n to x) p^x(1-p)^n-x
Nelson
why it is not possible brother
ibrar
were n= 2 ( binomial) x= number of makes (0 to 6) and p= probability, could be 0.8.
Nelson
so after we calculate the expected values for each observed value (f) we do the chi-square. x^2=summatory(observed-expected)^2 / expected
and compare with x^2 in table with 0.8
Nelson
tomorrow I'll post the answer, I'm so tired today, sorry for my mistake in the first messages.
Nelson
It is possible, sorry for my mistake
Nelson
two trader shared investment and buoght Cattle.Mr.Omer bought 255 cows & rented the farm for a period of 32 days. Mr. Ahmed grazed his Cattle for 25 days. Mr. Ahmed's cattle was 180 cows.Together they profited $ 7800. the rent of the farm is $ 3000 so divide the profit per gows/day for grazing day
Mohamed
how to start this book, who is reading thins first time
This is hard to type, so I'll use "m" for "x bar", and a few other notations that I hope will be clear: Definition: sqrt(SUM[(x - m)^2] / (n-1)) where m = SUM[x] / n Desired formula: sqrt((SUM[x^2] - SUM[x]^2)/n / (n-1)) Now let's do what you started to do, and see if we can manipulate the definitio
Michael
what is the difference between (n ) and (n-1) in the mean and variance
Soran
Definition: sqrt(SUM[(x - m)^2] / (n-1)) where m = SUM[x] / n
what is the difference between (n and n-1)
Soran
Hi, the diference is tha when we estimate parameters in a sample (not in the total population) we need to consider the degrees of liberty for the estimation.
Nelson
Hie guys, am analysing rainfall data for different stations and i got kurtosis values of 0.7 for one station and 0.4 for another, what can i say about this?
I want to ask, as a student who wants to study mechatronic engineering in the University. What are the necessary subject he or she needed, and after ur service were can or she work as a graduate
A hypothesis in a scientific context, is a testable statement about the relationship between two or more variables or a proposed explanation for some observed phenomenon. In a scientific experiment or study, the hypothesis is a brief summation of the researcher's prediction of the study's findings.
Hamzah
Which may be supported or not by the outcome. Hypothesis testing is the core of the scientific method.
Hamzah
statistics means interpretation analysis and representation of numerical data
Ramzan
To check the statment or assumption about population parameter is xalled hypothesis