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What is the probability that the birthdays of five persons chosen at random will fall in twelve different calendar months?
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∙ 14y ago
The probability that the birthdays of five persons chosen at random will fall in twelve different calender months is zero.
You would need at least twelve persons to have a non zero probability.
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∙ 14y ago
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Those influences in a persons life that are believed to increase the probability that an individual will engage in criminal behavior are called?
Risk factors
If the probability of getting identical twins is 0.004 while the probability of fraternal twins is 0.023 then what is the probability that a person chosen will be a twin?
P(twin)=P(identical)+P(fraternal) P(twin)=0.004 +0.023 P(twin)=0.027 [That's 27 out of 1,000 cases, or 54 persons out of 1027.]
What is the probability that in a room of 8 people 2 have the same birthday?
P = 0.072314699... ≈ 7.23%SOLUTIONLet's identify the 8 people by integers 1, 2,..., 8, and their birthday by x1, x2,..., x8.To include the possibility of xi = Feb. 29 (of a leap year), we will consider 4(365) +1 = 1451 days, so the probability of a person having Feb. 29 as his birthday isP(xi = Feb. 29) = 1/1461. The probability of a person having his birthday on anygiven day except Feb. 29 is P(xi ≠ Feb. 29) = 4/1461.Let's first calculate the probability of person 1 and 2 having the same birthday, x1 = x2, and persons 3, 4,..., 8, all have different birthdays x3,...,x8.Now we go.We take person 1 and have two possibilities:A).- x1 = Feb. 29 B).- x1 ≠ Feb. 29P(x1 = Feb. 29) = 1/1461; P(x ≠ Feb. 29) = 1 - 1/1461We will construct two branches. One for each possibility.Branch A: given x1 = Feb. 29, the probability of person 2 having the same birthdayis P(x2 = x1) = 1 - 1/1461. The probability of person 3 not having the same birthday is P(x3 ≠ x1 = x2) = 1 - 1/1461. The probability of person 4 not having either two birthdays is P(x4 ≠ x1, x4 ≠ x3) = 1 - 5/1461. The probability of person 5 not sharing x1 nor x3 nor x4 is P(x5 ≠ x4, x5 ≠ x3, x5 ≠ x2) = 1 - 9/1461. And so a series of terms develops:P(xi ≠ ... ) = 1 - (4i - 3)/1461 from i = 1, to i = n-2where n is the number of people in the room (In this particular question for 8people, n = 8).The branch we have constructed for possibility A is the product of the individualconditional probabilities;A: (1/1461)2 π1n-2[1-(4i-3)/1461]where π1n-2 is the product operator "pi" of "i" terms from i =1, to i = n-2.Branch B: Given x1 ≠Feb. 29, the probability of person 2 having the same birthdayis P(x2 = x2) = 4/1461. The probability of person 3 not having the same birthdayis P(x3 ≠ x1) = 1 - 4/1461. The probability of person 4 not sharing birthdays with 2 and 3 is P(x4 ≠ x3, x4 ≠ x2) = 1 - 8/1451. The probability of person 5 notsharing x2 nor x3 nor x4 is P(x5 ≠ ...) = 1 - 12/1461. And so a series of termsdevelops:P(xi ≠ ... ) = 1- (4i -4)/1461 from i = 2, to i = n-1where n is the number of people in the room (In this particular question for 8 people, n = 8).Branch for possibility B: (1-1/1461)(4/1461) π2n-1[1-(4i-4)/1461]The sum of these two branch expressions give the probability of persons 1 and 2sharing a birthday in a group of n persons:P(x1=x2)=(1/1451)2π1n-2[1-(4i-3)/1461]+(1-1/1461)(4/1461)π2n-1[1-(4i-4)/1461]Now, the probability of any two persons, and only those two persons, to share a birthday in a group of n persons, is the previous probability, times the number of different combinations of two persons, (i,j), in a group of n persons, nC2(where, nC2 = n!/[2!(n-2)!] )The final expression is:--------------------------------------------------------------------------------------------P = nC2{(1/1451)2π1n-2[1-(4i-3)/1461]+(1-1/1461)(4/1461)π2n-1[1-(4i-4)/1461]}--------------------------------------------------------------------------------------------For the case of n = 8, the result is, P = 0.072314699... ≈ 7.23%
How many different 5 persons committees can be formed from a group of 8 women and 6 men with each committee containing at least 3 women?
1316
10 persons can watch a movie in 7 days What is the probability that at least two them watch on the same day?
Not sure but could it be:- Probability of selecting one day out of seven:-(1/7) Now probability that only one person out of 10 watches the movie:-(1/10) Thus the total probability of only one person watching the movie on any given day will be:- (1/7)*(1/10)=1/70 Now, the probability that at least two of them watch the movie on the same day will be:- 1-(1/70)=(69/70) Hope this works, Any other suggestions are welcome. - Regards, NightHawk
What is probability of two persons being born on different days?
364 out of 365
What is the probability that 4 persons can sit in 4 different chairs?
4 x 4 = 16
What is the probability that a single randomly selected juror from Hidalgo County is Mexican-American?
If X = Mexican-American persons eligible for jury service in Hidalgo CountyIf Y = All persons eligible for jury service in Hidalgo Countythen the probability is X/Y.
What approach to probability is based on a persons degree of belief and hunches that a particular event will happen?
Subjective If you assume particular events will happen with a certain prior distribution, that is Bayesian probability.
What has the author Alex Riddell written?
Alex Riddell has written: 'In old Belfast' 'The Ulster calendar of persons and events'
Those influences in a persons life that are believed to increase the probability that an individual will engage in criminal behavior are called?
Risk factors
What is probability that 2 persons sharing same birth date?
If you assume that birth dates are uniformly distributed over the year (they are not), and you ignore leap years, then the probability of two people selected at random, share a birthday is 1/365.
If the probability of getting identical twins is 0.004 while the probability of fraternal twins is 0.023 then what is the probability that a person chosen will be a twin?
P(twin)=P(identical)+P(fraternal) P(twin)=0.004 +0.023 P(twin)=0.027 [That's 27 out of 1,000 cases, or 54 persons out of 1027.]
Different persons have different DNA fingerprints because?
they have different numbers of the same VNTR.
What is the probability that in a room of 8 people 2 have the same birthday?
P = 0.072314699... ≈ 7.23%SOLUTIONLet's identify the 8 people by integers 1, 2,..., 8, and their birthday by x1, x2,..., x8.To include the possibility of xi = Feb. 29 (of a leap year), we will consider 4(365) +1 = 1451 days, so the probability of a person having Feb. 29 as his birthday isP(xi = Feb. 29) = 1/1461. The probability of a person having his birthday on anygiven day except Feb. 29 is P(xi ≠ Feb. 29) = 4/1461.Let's first calculate the probability of person 1 and 2 having the same birthday, x1 = x2, and persons 3, 4,..., 8, all have different birthdays x3,...,x8.Now we go.We take person 1 and have two possibilities:A).- x1 = Feb. 29 B).- x1 ≠ Feb. 29P(x1 = Feb. 29) = 1/1461; P(x ≠ Feb. 29) = 1 - 1/1461We will construct two branches. One for each possibility.Branch A: given x1 = Feb. 29, the probability of person 2 having the same birthdayis P(x2 = x1) = 1 - 1/1461. The probability of person 3 not having the same birthday is P(x3 ≠ x1 = x2) = 1 - 1/1461. The probability of person 4 not having either two birthdays is P(x4 ≠ x1, x4 ≠ x3) = 1 - 5/1461. The probability of person 5 not sharing x1 nor x3 nor x4 is P(x5 ≠ x4, x5 ≠ x3, x5 ≠ x2) = 1 - 9/1461. And so a series of terms develops:P(xi ≠ ... ) = 1 - (4i - 3)/1461 from i = 1, to i = n-2where n is the number of people in the room (In this particular question for 8people, n = 8).The branch we have constructed for possibility A is the product of the individualconditional probabilities;A: (1/1461)2 π1n-2[1-(4i-3)/1461]where π1n-2 is the product operator "pi" of "i" terms from i =1, to i = n-2.Branch B: Given x1 ≠Feb. 29, the probability of person 2 having the same birthdayis P(x2 = x2) = 4/1461. The probability of person 3 not having the same birthdayis P(x3 ≠ x1) = 1 - 4/1461. The probability of person 4 not sharing birthdays with 2 and 3 is P(x4 ≠ x3, x4 ≠ x2) = 1 - 8/1451. The probability of person 5 notsharing x2 nor x3 nor x4 is P(x5 ≠ ...) = 1 - 12/1461. And so a series of termsdevelops:P(xi ≠ ... ) = 1- (4i -4)/1461 from i = 2, to i = n-1where n is the number of people in the room (In this particular question for 8 people, n = 8).Branch for possibility B: (1-1/1461)(4/1461) π2n-1[1-(4i-4)/1461]The sum of these two branch expressions give the probability of persons 1 and 2sharing a birthday in a group of n persons:P(x1=x2)=(1/1451)2π1n-2[1-(4i-3)/1461]+(1-1/1461)(4/1461)π2n-1[1-(4i-4)/1461]Now, the probability of any two persons, and only those two persons, to share a birthday in a group of n persons, is the previous probability, times the number of different combinations of two persons, (i,j), in a group of n persons, nC2(where, nC2 = n!/[2!(n-2)!] )The final expression is:--------------------------------------------------------------------------------------------P = nC2{(1/1451)2π1n-2[1-(4i-3)/1461]+(1-1/1461)(4/1461)π2n-1[1-(4i-4)/1461]}--------------------------------------------------------------------------------------------For the case of n = 8, the result is, P = 0.072314699... ≈ 7.23%
What is persons heart rate per minute?
Every persons heartrate is different! Depending on your height and weight.
What is the probabililty of at least 2 people same birthday from a group of 13 people?
19.4%CALCULATION:The probability of at least 2 people having the same birthday in a group of 13people is equal to one minus the probability of non of the 13 people having thesame birthday.Now, lets estimate the probability of non of the 13 people having the same birthday.(We will not consider 'leap year' for simplicity, plus it's effect on result is minimum)1. We select the 1st person. Good!.2. We select the 2nd person. The probability that he doesn't share the samebirthday with the 1st person is: 364/365.3. We select the 3rd person. The probability that he doesn't share the samebirthday with 1st and 2nd persons given that the 1st and 2nd don't share the samebirthday is: 363/365.4. And so forth until we select the 13th person. The probability that he doesn'tshare birthday with the previous 12 persons given that they also don't sharebirthdays among them is: 353/365.5. Then the probability that non of the 13 people share birthdays is:P(non of 13 share bd) = (364/365)(363/365)(362/365)∙∙∙(354/365)(353/365)P(non of 13 share bd) ≈ 0.805589724...Finally, the probability that at least 2 people share a birthday in a group of 13people is ≈ 1 - 0.80558... ≈ 0.194 ≈ 19.4%The above expression can be generalized to give the probability of at least x =2people sharing a birthday in a group of n people as:P(x≥2,n) = 1 - (1/365)n [365!/(365-n)!]
