Tutors Answer Your Questions about Probability-and-statistics (FREE)
Question 21656: I dont know where to begin with this, please help me!
After you complete this exam, you and two friends decide to go to McDonald’s to celebrate. Last month McDonald’s filled 90% of the orders accurately. What is the probability that:
a) All three orders will be filled accurately?
b) None of the three orders will be filled accurately?
c) At least 2 of the 3 orders will be filled accurately?
Click here to see answer by longjonsilver(2297)  |
Question 21888: The following data is for worker output in a plant, for different hours of the day, with a numerical measurement on their production:
hour - 1 2 3 4 5 6 7
output - .5 2.6 3.6 4.5 4.9 5.4 5.7
Find the natural logarithmic regression model for this data
Click here to see answer by AnlytcPhil(1276)  |
Question 22696: A shipment of light bulbs undergoes an inspection where 200 bubls are testes and 6 turn out defective.
a.What is the probability that a light bulb in this shipment is defective?
b. If the shipment contains 10280 light bulbs, predict how many from the entire shipment will be defective.
Click here to see answer by rapaljer(4667)  |
Question 21668: Consider the experiment of rolling two fair dice.
a. What is the probability that you get either: "snake eyes" of "double boxcars"?
b. What is the probability that the total number of points shown on these two dice is a perfect square?
c. What is the probability that the total number of points shown on these two dice is a prime number?
Click here to see answer by venugopalramana(3286) |
Question 21547: Please help me! i have no clue where to even begin!
A statistical analysis of 1,000 long distance telephone calls made from the headquarters of the Bricks and Clicks Computer Corporation indicates that the length of these calls is normally distributed with m = 240 seconds and s = 40 seconds.
a) What is the probability that a call lasts less than180 seconds?
b) What is the probability that a particular call lasts between 180 and 300 seconds?
c) What is the length of a call if only 1% of all calls are shorter?
Click here to see answer by venugopalramana(3286) |
Question 22312: You get a box of assorted chocolates for Christmas. There are the following
types; 8 dark chocolate-coated caramels, 7 milks chocolate-coated caramels,7
dark chocolate-coated nuts, 6 milk chocolate-coated vanilla creams, 8 dark
chocolate-coated cherries, 7 milk chocolate-coated strawberry creams. Determine
the following probabilitie;
of getting 2 dark chocolate-coated cherries in a row and then 5
caramels in a row.''.
Click here to see answer by venugopalramana(3286) |
Question 21657: Here's another that i think i know how to do, but am not really sure. please help me out!!!
A statistical analysis of 1,000 long distance telephone calls made from the headquarters of the Bricks and Clicks Computer Corporation indicates that the length of these calls is normally distributed with m = 240 seconds and s = 40 seconds.
a) What is the probability that a call lasts less than180 seconds?
b) What is the probability that a particular call lasts between 180 and 300 seconds?
c) What is the length of a call if only 1% of all calls are shorter?
Click here to see answer by venugopalramana(3286) |
Question 23954: ADDED ADDITIONAL EXPLANATION AND WORKING FOR THE SECOND PROBLEM..ONLY YOU NEED TO DO THE ARITHMATIC
VENUGOPAL
OH!THIS IS A RESEARCH PROJECT FOR YOU ..THEN IT IS OK ..
THEN LET US ANALYSE A BIT..THE FIRST PART IS EASY .I THINK YOU KNOW IT AND DID NOT ASK FOR THE ANSWER.BUT LET US BEGIN WITH THAT .IT WILL SERVE TO OUT LINE THE PROCEDURE AND TO ATTEMPT THE SECOND PROBLEM..
it says that the probability of 2 out of 30 people having the same birthday is 70 percent.
IN THE CONVENTIONAL EASY WAY FOR HIGH SCHOOL STUDENTS UPTO K-12 STANDARD,THIS IS DONE AS FOLLOWS WITH 365 DAYS ONLY CONSIDERED PER YEAR & 30 NUMBER GROUP.
PROBABILITY OF 2 HAVING SAME BIRTH DAY IN THE GROUP = 1 - HAVING NO COINCIDENCE IN BIRTH DAY .....
FOR NO COINCIDENCE TO OCCURR ...............
THE FIRST PERSON CAN TAKE BIRTH ON ANY OF THE 365 DAYS,
THE SECOND PERSON'S BIRTH DAY CAN BE ONLY FROM THE REST 365-1=364 DAYS
THE THIRD PERSON'S BIRTH DAY CAN BE ONLY FROM THE REST 365-2=363 DAYS
....................ETC.....
THE THIRTIETH PERSON'S BIRTH DAY CAN BE ONLY FROM THE REST 365-29=336 DAYS
SO NUMBER OF ALL POSSIBILITIES FOR NO COINCIDENCE =365*364*363*.......*336
TOTAL NUMBER OF POSSIBILITIES WITH NO RESTRICTIONS=365*365*365....30 TIMES =365^30
HENCE PROBABILITY OF NO COINCIDENCE =365*364*363*.......*336/365^30=0.293683757
HENCE PROBABILITY FOR A COINCIDENCE =1- 0.293683757 = 0.706316243..OR ABOUT 70% AS YOU HAVE GOT.
HOWEVER THIS METHOD CAN NOT BE USED FOR 3 OR 4 OR 6 COINCIDENCES AS YOU WANTED FOR SECOND CASE ...
SO LET US EVOLVE A METHOD SUITABLE FOR THAT GENERAL CASE...
I HOPE YOU KNOW PIGEON HOLE PRICIPLE WHICH IS A SIMPLE BUT EFFECTIVE TOOL FOR EXISTENCE PROBLEMS.....
IN SIMPLE TERMS,IT SAYS THAT IF THERE ARE 10 PIGEONS AND 9 CAGES THERE WILL BE ATLEAST ONE CAGE WITH MORE THAN ONE PIGEON WHICH IS OBVIOUSLY CORRECT .
WE WONT USE THE PRINCIPLE DIRECTLY BUT USE ITS ANALOGY TO SOLVE OUR PROBLEM.
SO NOW LET US THINK OF OUR 365 DAYS AS 365 CAGES.30 PERSONS AS 30 PIGEONS.
SO WE WANT OUR CAGES TO BE FILLED AS FOLLOWS FOR OUR FIRST PROBLEM OF ONE COINCIDENCE FOR WHICH WE GOT THE ANSWER OF 70 % NOW.
LET US PUT OUR REQUIREMENT AS NUMBER OF 30 PIGEONS (NUMBER OF PERSONS=30)IN EACH OF 365 CAGES (NUMBER OF DAYS IN AN YEAR = 365 )
here days are different cages and persons are pigeons.
A ...first is CBD-common birthday type cages.
A=[2].TO SHOW THAT WE WANT ONE CAGE WITH 2 PERSONS HAVING SAME BIRTH DAY.
B....second is DBD-different birth day type cages.
B=[1,1,......28 NOS.]THAT IS 28 CAGES WITH 1 PERSON EACH HAVING DIFFERENT BIRTH DAYS OUT OF THE REST 365-1=364 DAYS.
C....third is rest or empty cages.
C=[0,0,....365-30=335 NOS.]THAT IS 335 CAGES WITH NO PERSONS .
THAT IS figures in square brackets [ ] show number of persons in each of different types of cages mentioned above.
[2] indicates 2 persons are there in CBD type.if there are 6 persons with same birth day it will be shown as [6].but,if the 6 are in 3 pairs each with a different common birth day it will be shown as [2,2,2].
LET US SUM IT UP AS DIFFERENT TYPES OF CAGES AS EXPLAINED ABOVE....
A = (1),B =(28),C =(335)..THAT IS 1 CAGE WITH 2 BOYS HAVING SAME BIRTH DAY ,28 CAGES WITH EACH PERSON HAVING DIFFERENT BIRTH DAYS AND 335 CAGES WITH NO PERSONS.
THAT IS figures in parethesis ( ) show the number of types of cages in each catype explained above.(1)under A shows one cage of CBD type.(21)under B indicates 21 cages of DBD type...etc...
we calculate first the ways of selecting the cages (365 days in the year) in to the 3 types ,that is common birth day to be say 1st.jan.,different birthdays to be 2nd.,jan to 30th.jan...…etc...then we find ,the pigeons (30 persons )to be selected to fit those days .we multiply the two to find total number of arrangements possible. we divide it with total of all arrangements without any condition = 365^30 to get the probability.
WITH THIS DISPENSATION WE CAN CALCULATE EACH COMBINATION USING THE FOLLOWING STANDARD FORMULAE..
NUMBER OF COMBINATION OF DIVIDING N THINGS IN TO K GROUPS EACH HAVING SAY X,Y,Z ETC...THINGS SUCH THAT X+Y+Z...=K...IS GIVEN BY N!/X!*Y!*Z!*.....
SO NOW LET US APPLY THIS TO OUR FIRST PROBLEM AND THEN THE SECOND PROBLEM...
*************************************
FIRST PROBLEM.....PROBABILITY OF ATLEAST ONE COINCIDENCE
AS EARLIER THIS IS = 1-PROBABILITY OF NO COINCIDENCE..
B=[1,1....30]
C=[0,0,....335]..OR
HERE A DOES NOT EXIST (0) AS THERE ARE NO COINCIDENCES B=(30),C=(335)..COMBINATION..
1. WE FIRST FIND NUMBER OF SUCH COMBINATIONS POSSIBLE THAT IS 30 OF 1 EACH TYPE AND 335 OF 0 EACH TYPE FROM A TOTAL OF 365....
=365!/(30!*335!)=G SAY
2.NEXT WE FIND THE NUMBER OF POSSIBLE ARRANGEMENTS WITHIN EACH COMBINATION. B= 30!/1!*1!*1!....=30!...
HENCE NO COINCIDENCE COMBINATIONS BY MULTIPLICATION THEOREM IS
#NAME?
PROBABILITY OF NO COINCIDENCE COMBINATIONS =G*B/(365^30)
=365!*30!/(30!*335!*365^30)
=365!/(335!*365^30)=0.293683757
PROBABILITY OF ATLEAST 1 COINCIDENCE =1-G*B/(365^30)=1-0.293683757=0.706316243
OR ABOUT 70 %.=PROBABILITY OF 2 OR MORE HAVING SAME BIRTH DAY.
**********************************************
FIRST PROBLEM WITH A CHANGE .THAT THERE SHOULD BE ONLY 1 COINCIDENCE OR ONLY 2 PEOPLE WILL HAVE SAME BIRTH DAY.USING SAME NOTATION AS ABOVE...
A=[2],B[1,1,....28 CAGES],[0,0,....335 CAGES]
OR...A=(1),B=(28),C=(336)
G=365!/(1!*28!*336!)
A*B=30!/(2!...1 TIME ONLY )*(1!*1!*1!...28 TIMES )=30!/2!
PROBABILITY OF ONLY ONE COINCIDENCE=G*A*B/365^30
=365!*30!/(1!*28!*336!*2!*365^30)
=0.380215027..OR ABOUT 38%
************************************************
NOW THE 6 HAVING SAME BIRTH DAY CAlCULATION
A=[6],B=[1,1,....24 CAGES],C=[0,0,....335 CAGES]
OR...A=(1),B=(24),C=(340)
G=365!/(1!*24!*340!)
A*B=30!/(6!)
PROBABILITY OF 6 COINCIDENCES =(365!*30!)/(1!*24!*340!*6!*365^30)=
Hello,
I have been researching the birthday paradox and it says that the probability of 2 out of 30 people having the same birthday is 70 percent. Then it goes on to say that the probability of 6 out of 30 people having the same birthday is 12 percent. My question is what formulas would I use to find the probablity of 6 out of 30 is 12 percent? This seems too high. Is it even right?
thanks
pierre
Click here to see answer by venugopalramana(3286) |
Question 24752: three different airlines fly from bowling green to lexington. those same three airlines and two other airlines fly from lexington to louisville. there are no direct flights from bowling green to louisville..
How many ways can a traveler book flights from bowling green to lousivelle?
Click here to see answer by stanbon(57251) |
Question 24776: according to the national center for health statistics, each time a baby is born the probablility that it is a boy is always 51.3% and a girl is 48.7% regardless of the sex of any previous births in the family. suppose a couple has four children. what is the probability they are all grils?? what is the probability there are one gril and three boys
Click here to see answer by josmiceli(9661)  |
Question 25644: each day the school cafeteria serves 3 sandwiches,4 diffrent drinks,and 5 diffrent types of fruits.if you pick randomly,what is the probability that you will not get a tuna sandwich,an orange juice,and an apple.
Click here to see answer by stanbon(57251) |
Question 26843: Don't get this one :(
A bag contains 6 red marbles, 9 blue marbles, and 5 green marbles. You withdraw one marble, replace it, and then withdraw another marble. What is the probability that you do NOT pick two green marbles?
Click here to see answer by AnlytcPhil(1276)  |
Question 28525: hi
i need help on this question: five cards are drawn from an ordinary deck of 52 playing cards. what is the probability that the hand drawn is a full house (two of one kind, three of another kind)? I don't even know how to start this problem. Please help!
Click here to see answer by venugopalramana(3286) |
Question 28645: hi
i need help with this problem. i need help on this question: five cards are drawn from an ordinary deck of 52 playing cards. what is the probability that the hand drawn is a full house? (a full house means three of one kind, three of another like three queens and two nines, etc.) please help!
Click here to see answer by stanbon(57251) |
Question 29565: A deck of cards has 6 gray, 5 navy, and 6 yellow cards. You pick 4 cards from the deck. Cards are not returned to the deck after they are picked. P(the first card is not gray, the second card is gray, the third card is yellow, and the fourth card is gray).
Click here to see answer by longjonsilver(2297)  |
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