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
statistics is the beach of mathematics which deals with collection ,organisation, presentation, analysis and interpretation of numerical data
Saeed
oh but interpretation of data, like what and how? 🤔
Bhavani
interpretation: Think in a way that you have given a company year turnover and you have a record of 100years and data set is like (Year,Turnover). Now with that data you can interpret many thing how was the company growth, when were the losses and other things
Akash
interpretation: it is a process in which we make a decision about a population on the basis of sample data .
example: if we want to interpret the average income of employees for upcoming year so we have to interpret the income of employees on the basis of previous year's income of those employees
I would say art is a creation. A chef is an artist. They create new dishes just like the painters. I believe one who creates something new, is an artist. So, Statistics is also an art, if you know it, you can create some new formula, theory, law, etcetera. It is also Science. So yes, it is both.
Rohan
how do you use the normal distribution table when testing the hypothesis
Davia
I am sorry Davia, I cannot help you with that.
Rohan
percentages of all the possible outcomes are measured. This is so simple and bases on the questionnaire or interview schedule. It's just measuring the probability chances of high %age of the either part of the hypothesis ... dependent ..independent.
data is classified on the basis of respondents
saifuddin
what percent of the students would be expected to score above 95?
Mean (average) 4...
Median (middle term) 3.5..
Mode (frequency) every element in a set has 1 frequrncy
Akash
i arrange the data set in ascending order. that is, 1,2,3,4,5,6. then find the data set that falls in the middle. in this case, 3 & 4 fall in the middle. you then sum and obtain the average. that is, (3+4)/2=3.5. therefore, 3.5 is the median.
Gbenga
both of you are correct.
Joseph
hello guys
Abasikponke
thanks
lucy
great to be here
King
how does a line graph look
King
hi
Davia
hello
lucy
pls who knows how line graph look like
King
line graph usually have a straight line running through axis
Dike
am new here anyone willing to orient me?
Timothy
find the media of the following numbers 61,64,67,70,73
what is the percentile for the set of data in the class C and frequency F(c,f)given by (9.3-9.7,2) (9.8-10.2,5) (10.3-10.7,12) (10.8-11.2,17) (11.3-11.7,14) (11.8-12.2,6) (12.3-12.7,3) (12.8-13.2,1)
arrange ascending and desending order than the mid value is Median
rajendra
ok
Hrishe
what if it is a group data
Oloyede
mean/ medium/ mode
Michelle
n\2 and n+1\2
asad
An operational manager at a manufacturing company is interested in the level of satisfaction of computer buyers. The manager has developed a satisfaction scale of 1-10 to mark their level of understanding with the company.What is the population of the interest?
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