Respuesta :
Answer:
316
Step-by-step explanation:
The question has a little problem:
"the probability that they chose the same book is m n for relatively prime positive integers m and n. Compute 100m + n."
The correct sentence:
the probability that they chose the same book is "m/n" for relatively prime positive integers m and n.
Total number of books = 4
We have:
Number of 200page book = 1
Number of 400page book = 1
Number of 600page book = 1
Number of 800page book = 1
Probability of picking same book:
Velma read page 122 of her book Daphne read page 304 of her book
If it is same book, it must contain atleast 400page.
Therefore, 400page, 600page and 800 page would be considered in the probability.
Pr(one 400page) = 1/4
Pr(picking two 400page) = 1/4 * 1/4 = 1/16
Pr(one 600page) = 1/4
Pr(picking two 600page) = 1/4 * 1/4 = 1/16
Pr(one 800page) = 1/4
Pr(picking two 800page) = 1/4 * 1/4 = 1/16
Pr(picking same book page)=
Pr(picking two 400page) or Pr(picking two 600page) or Pr(picking two 800page)
= Pr(picking two 400page) + Pr(picking two 600page) + Pr(picking two 800page) = 1/16+ 1/16+ 1/16
Pr(picking same book page)= 3/16
This answer satisfies the probability as m/n for relatively prime positive integers m and n.
Two numbers are said to be relatively prime integers if the only positive integer that divides both of them is 1. It means the numerator and denominator of the fraction have been reduced to the lowest form.
m/n = 3/16
m = 3, n= 16
100m + n = 100(3) + 16
= 316
Answer:
100m+n = 6425
Step-by-step explanation:
Let X be the book Velma picks and Y the book that Daphne picks. Note that X and Y are independent and identically distributed, so for computations, i will just focus on X for now.
Lets denote with A, B, C and D the books with 200, 400, 600 and 800 pages respectively.
Note that, without any restriction P(X=A) = P(X=B) = P(X=C) = P(X=D) = 1/4. However, if we also add the condition that R = 122, where R is the page picked, we will need to apply the Bayes formula. For example,[tex]P(X=A|R=122) = \frac{P(R = 122|X=A)*P(X=A)}{P(R=122)}[/tex]
P(X=A) is 1/4 as we know, and the probability P(R=122|X=A) is basically the probability of pick a specific page from the book of 200 pages long, which is 1/200 (Note however, that if he had that R were greater than 200, then the result would be 0).
We still need to compute P(R=122), which will be needed in every conditional probability we will calcultate. In order to compute P(R=122) we will use the Theorem of Total Probability, in other words, we will divide the event R=122 in disjoint conditions cover all possible putcomes. In this case, we will divide on wheather X=A, X=B, X=C or X=D.
[tex]P(R=122) = P(R=122 | X=A)*P(X=A) + P(R=122|X=B)*P(X=B)+P(R=122|X=C)*P(X=C)+P(R=122|X=D)*P(X=D) = 1/200 * 1/4 + 1/400*1/4 + 1/600*1/4+1/800*1/4 = 1/4*(12/2400 + 6/2400 + 4/2400 + 3/2400) = 1/384[/tex]
Thus, P(R=122) = 1/396
With this in mind, we obtain that
[tex]P(X=A|R=122) = \frac{\frac{1}{200}*\frac{1}{4}}{\frac{1}{396}} = \frac{384}{800} = \frac{12}{25}[/tex]
In a similar way, we can calculate the different values that X can take given that R = 122. The computation is exactly the same except that for example P(R=122|X=B), is 1/400 and not 1/200 because B has 400 pages.
[tex]P(X=B|R=122) = \frac{\frac{1}{400}*\frac{1}{4}}{\frac{1}{384}} = \frac{384}{1600} = \frac{6}{25}[/tex]
[tex]P(X=C|R=122) = \frac{\frac{1}{600}*\frac{1}{4}}{\frac{1}{384}} = \frac{384}{2400} = \frac{4}{25}[/tex]
[tex]P(X=D|R=122) = \frac{\frac{1}{800}*\frac{1}{4}}{\frac{1}{384}} = \frac{384}{3200} = \frac{3}{25}[/tex]
We can make the same computations to calculate the probability of Y = A,B,C or D, given that R=304. However, P(Y=A|R=304) will be 0 because A only has 200 pages (similarly, P(R=304|Y=A) = 0, R=304 and Y=A are not compatible events). First, lets compute the probability that R is 304.
[tex]P(R=304) = P(R=304|Y=A)*P(Y=A)+P(R=304|Y=B)*P(Y=B)+P(R=304|Y=C)*P(Y=C)+P(R=304|Y=D)*P(Y=D) = 0+1/400*1/4+1/600*1/4+1/800*1/4 = 13/9600[/tex]
Thus, P(R=304) = 13/9600. Now, lets compute each of the conditional probabilities
[tex] P(Y=A|R=304) = 0[/tex] (as we stated before)
[tex]P(Y=B|R=304) = \frac{\frac{1}{400}*\frac{1}{4}}{\frac{13}{9600}} = \frac{9600}{1600*13} = \frac{6}{13}[/tex]
[tex]P(Y=C|R=304) = \frac{\frac{1}{600}*\frac{1}{4}}{\frac{13}{9600}} = \frac{9600}{2400*13} = \frac{4}{13}[/tex]
[tex]P(Y=D|R=304) = \frac{\frac{1}{800}*\frac{1}{4}}{\frac{13}{9600}} = \frac{9600}{3200*13} = \frac{3}{13}[/tex]
We want P(X=Y) given that [tex] R_x = 122 [/tex] and [tex] R_y = 304 [/tex] (we put a subindex to specify which R goes to each variable). We will remove the conditionals to ease computations, but keep in mind that we are using them. For X to be equal to Y there are 3 possibilities: X=Y=B, X=Y=C and X=Y=D (remember that Y cant be A given that [tex] R_y = 304). Using independence, we can split the probability into a multiplication.
[tex] P(X=Y=B) = P(X=B|R=122)*P(Y=B|R=304) = \frac{6}{25} * \frac{6}{13} = \frac{36}{325} [/tex]
[tex] P(X=Y=C) = P(X=C|R=122)*P(Y=C|R=304) = \frac{4}{25}*\frac{4}{13} = \frac{16}{325} [/tex]
[tex] P(X=Y=D) = P(X=D|R=122)*P(Y=D|R=304) = \frac{3}{25}*\frac{3}{13} = \frac{9}{325} [/tex]
Therefore
[tex] P(X=Y) = \frac{36}{325} + \frac{16}{325} + \frac{9}{325} = \frac{61}{325} [/tex]
61 is prime and 325 = 25*13, thus, they are coprime. Therefore, we conclude that m = 61, n = 325, and thus, 100m+n = 6425.
