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,![P(X=A|R=122) = \frac{P(R = 122|X=A)*P(X=A)}{P(R=122)}](https://tex.z-dn.net/?f=P%28X%3DA%7CR%3D122%29%20%3D%20%5Cfrac%7BP%28R%20%3D%20122%7CX%3DA%29%2AP%28X%3DA%29%7D%7BP%28R%3D122%29%7D)
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.
Thus, P(R=122) = 1/396
With this in mind, we obtain that
![P(X=A|R=122) = \frac{\frac{1}{200}*\frac{1}{4}}{\frac{1}{396}} = \frac{384}{800} = \frac{12}{25}](https://tex.z-dn.net/?f=P%28X%3DA%7CR%3D122%29%20%3D%20%5Cfrac%7B%5Cfrac%7B1%7D%7B200%7D%2A%5Cfrac%7B1%7D%7B4%7D%7D%7B%5Cfrac%7B1%7D%7B396%7D%7D%20%3D%20%5Cfrac%7B384%7D%7B800%7D%20%3D%20%5Cfrac%7B12%7D%7B25%7D)
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.
![P(X=B|R=122) = \frac{\frac{1}{400}*\frac{1}{4}}{\frac{1}{384}} = \frac{384}{1600} = \frac{6}{25}](https://tex.z-dn.net/?f=P%28X%3DB%7CR%3D122%29%20%3D%20%5Cfrac%7B%5Cfrac%7B1%7D%7B400%7D%2A%5Cfrac%7B1%7D%7B4%7D%7D%7B%5Cfrac%7B1%7D%7B384%7D%7D%20%3D%20%5Cfrac%7B384%7D%7B1600%7D%20%3D%20%5Cfrac%7B6%7D%7B25%7D)
![P(X=C|R=122) = \frac{\frac{1}{600}*\frac{1}{4}}{\frac{1}{384}} = \frac{384}{2400} = \frac{4}{25}](https://tex.z-dn.net/?f=P%28X%3DC%7CR%3D122%29%20%3D%20%5Cfrac%7B%5Cfrac%7B1%7D%7B600%7D%2A%5Cfrac%7B1%7D%7B4%7D%7D%7B%5Cfrac%7B1%7D%7B384%7D%7D%20%3D%20%5Cfrac%7B384%7D%7B2400%7D%20%3D%20%5Cfrac%7B4%7D%7B25%7D)
![P(X=D|R=122) = \frac{\frac{1}{800}*\frac{1}{4}}{\frac{1}{384}} = \frac{384}{3200} = \frac{3}{25}](https://tex.z-dn.net/?f=P%28X%3DD%7CR%3D122%29%20%3D%20%5Cfrac%7B%5Cfrac%7B1%7D%7B800%7D%2A%5Cfrac%7B1%7D%7B4%7D%7D%7B%5Cfrac%7B1%7D%7B384%7D%7D%20%3D%20%5Cfrac%7B384%7D%7B3200%7D%20%3D%20%5Cfrac%7B3%7D%7B25%7D)
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.
![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](https://tex.z-dn.net/?f=P%28R%3D304%29%20%3D%20P%28R%3D304%7CY%3DA%29%2AP%28Y%3DA%29%2BP%28R%3D304%7CY%3DB%29%2AP%28Y%3DB%29%2BP%28R%3D304%7CY%3DC%29%2AP%28Y%3DC%29%2BP%28R%3D304%7CY%3DD%29%2AP%28Y%3DD%29%20%3D%200%2B1%2F400%2A1%2F4%2B1%2F600%2A1%2F4%2B1%2F800%2A1%2F4%20%3D%2013%2F9600)
Thus, P(R=304) = 13/9600. Now, lets compute each of the conditional probabilities
(as we stated before)
![P(Y=B|R=304) = \frac{\frac{1}{400}*\frac{1}{4}}{\frac{13}{9600}} = \frac{9600}{1600*13} = \frac{6}{13}](https://tex.z-dn.net/?f=P%28Y%3DB%7CR%3D304%29%20%3D%20%5Cfrac%7B%5Cfrac%7B1%7D%7B400%7D%2A%5Cfrac%7B1%7D%7B4%7D%7D%7B%5Cfrac%7B13%7D%7B9600%7D%7D%20%3D%20%5Cfrac%7B9600%7D%7B1600%2A13%7D%20%3D%20%5Cfrac%7B6%7D%7B13%7D)
![P(Y=C|R=304) = \frac{\frac{1}{600}*\frac{1}{4}}{\frac{13}{9600}} = \frac{9600}{2400*13} = \frac{4}{13}](https://tex.z-dn.net/?f=P%28Y%3DC%7CR%3D304%29%20%3D%20%5Cfrac%7B%5Cfrac%7B1%7D%7B600%7D%2A%5Cfrac%7B1%7D%7B4%7D%7D%7B%5Cfrac%7B13%7D%7B9600%7D%7D%20%3D%20%5Cfrac%7B9600%7D%7B2400%2A13%7D%20%3D%20%5Cfrac%7B4%7D%7B13%7D)
![P(Y=D|R=304) = \frac{\frac{1}{800}*\frac{1}{4}}{\frac{13}{9600}} = \frac{9600}{3200*13} = \frac{3}{13}](https://tex.z-dn.net/?f=P%28Y%3DD%7CR%3D304%29%20%3D%20%5Cfrac%7B%5Cfrac%7B1%7D%7B800%7D%2A%5Cfrac%7B1%7D%7B4%7D%7D%7B%5Cfrac%7B13%7D%7B9600%7D%7D%20%3D%20%5Cfrac%7B9600%7D%7B3200%2A13%7D%20%3D%20%5Cfrac%7B3%7D%7B13%7D)
We want P(X=Y) given that
and
(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 ![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}](https://tex.z-dn.net/?f=%20R_y%20%3D%20304%29.%20Using%20%3Cstrong%3Eindependence%3C%2Fstrong%3E%2C%20we%20can%20split%20the%20probability%20into%20a%20multiplication.%3C%2Fp%3E%3Cp%3E%5Btex%5D%20P%28X%3DY%3DB%29%20%3D%20P%28X%3DB%7CR%3D122%29%2AP%28Y%3DB%7CR%3D304%29%20%3D%20%5Cfrac%7B6%7D%7B25%7D%20%2A%20%5Cfrac%7B6%7D%7B13%7D%20%3D%20%5Cfrac%7B36%7D%7B325%7D%20)
![P(X=Y=C) = P(X=C|R=122)*P(Y=C|R=304) = \frac{4}{25}*\frac{4}{13} = \frac{16}{325}](https://tex.z-dn.net/?f=%20P%28X%3DY%3DC%29%20%3D%20P%28X%3DC%7CR%3D122%29%2AP%28Y%3DC%7CR%3D304%29%20%3D%20%5Cfrac%7B4%7D%7B25%7D%2A%5Cfrac%7B4%7D%7B13%7D%20%3D%20%5Cfrac%7B16%7D%7B325%7D%20)
![P(X=Y=D) = P(X=D|R=122)*P(Y=D|R=304) = \frac{3}{25}*\frac{3}{13} = \frac{9}{325}](https://tex.z-dn.net/?f=%20P%28X%3DY%3DD%29%20%3D%20P%28X%3DD%7CR%3D122%29%2AP%28Y%3DD%7CR%3D304%29%20%3D%20%5Cfrac%7B3%7D%7B25%7D%2A%5Cfrac%7B3%7D%7B13%7D%20%3D%20%5Cfrac%7B9%7D%7B325%7D%20)
Therefore
![P(X=Y) = \frac{36}{325} + \frac{16}{325} + \frac{9}{325} = \frac{61}{325}](https://tex.z-dn.net/?f=%20P%28X%3DY%29%20%3D%20%5Cfrac%7B36%7D%7B325%7D%20%2B%20%5Cfrac%7B16%7D%7B325%7D%20%2B%20%5Cfrac%7B9%7D%7B325%7D%20%3D%20%5Cfrac%7B61%7D%7B325%7D%20)
61 is prime and 325 = 25*13, thus, they are coprime. Therefore, we conclude that m = 61, n = 325, and thus, 100m+n = 6425.