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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%
If Two dice are rolled Find the probability of getting a sum greater than 8?
The first die can be from 2 to 6, it can not be one. So, the first die has a 5/6 chance. The second die must be a 2 if 6 is on the first die, 3 if 5 is on the first die, etc. so it has a 1/6 chance. Since the two events must occur ("and" situation) and they are independent: (1/6) * 5/6 = 5/36. A second way of thinking about this is, that each die can have values from 1 to 6, so you can make a x-y table (1-6 columns and 1-6 rows) and see 36 combinations, of which only 5 will add to 8: (2,6) (6,2) (3,5) (5,3) (4,4) so the answer is 5/36. We know that there are 36 different possible dice rolls. Since the dice must be greater than 8, we have the following combinations: 3-6 x2 4-5 x2 4-6 x2 5-5 5-6 x2 6-6 Which is 10, or 5/18. If you looking for > or = to 8, we must add: 2-6 x2 3-5 x2 4-4 to this list. Which is 15 => 5/12
How can you use the digit 8 four times to make 89?
88 + 8/888 + 8/8 = 89
What is the probability that exactly 2 have some kind of defect?
The probability is determined by the binomial distribution. We consider p = probability of defect, q = probability of not defect, n = sample size, and x= number of defects in sample, in this case x=2. We calculate the probability as P(X = x) = n!/[(n-x)! x!] pxqn-x If sample size = 10 and p = 0.1 then: P(x= 2) = 10!/(8!x2!)(0.1)2(0.9)8 = 0.1937 You can find more about the binomial distribution under Wikipedia. It is important also to note the assumptions when using this distribution. It must be a random sample and the probability of defects is known.
What is the probability of getting one head when flipping a coin three times?
The probability is 3/8.The probability is 3/8.The probability is 3/8.The probability is 3/8.
Factor x3 plus 8?
x3 + 8 = x3 + 23 = (x + 2)[x2 - (x)(2) + 22] = (x + 2) (x2 - 2x + 4)
Factor Completely x3 plus 8?
x3 + 8 = x3 + 23 = (x + 2)(x2 + 2x + 22) = (x + 2)(x2 + 2x + 4)
How do you factor x3-8x2?
x2(x - 8)
How do you factor X3-8?
(x+2)(x2+2x+4)
How do you factor x3-512?
(x - 8)(x2 + 8x + 64)
Factorise x cubed plus 8?
X3 + 8= (x+2)(x2-2x+4)
What is the factor of X3 - 2x2 - 4x plus 8?
x3 - 2x2 - 4x + 8 = (x2 - 4)(x - 2) = (x + 2)(x - 2)(x - 2)
Factor of 2x3 plus 16?
2x3 + 16 = 2(x3 + 8) = 2(x3 + 23) = 2(x + 2)(x2 - 2x + 22) = 2(x + 2)(x2 - 2x + 4)
What is the factorization of the trinomial x3 plus 13x2 plus 40x?
It is: (x2 + 5x) (x + 8)
how do i solve this 5x2 + 3x2 – x3 + 2x3 + 10x4 – 5x4?
x2 • (5x2 + x + 8)
Factor x3 plus 3x2-6x-8?
x3 + 3x2 - 6x - 8 = (x - 2)(x2 + 5x + 4) = (x - 2)(x + 1)(x + 4)
What is x to the third power subtracted by 5x to the second power plus 6x take away 8 factored?
x3 - 5x2 + 6x - 8 = x3 - 4x2 - x2 + 4x + 2x - 8 = x2(x - 4) - x(x - 4) + 2(x - 4) = (x - 4)(x2 - x + 2) which has no further real factors.
