## Conditional Probablity

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Fri Jan 27, 2012 10:13 pm
shloka19 says...

So this is a conditional probability question that is giving me a really hard time:
This is a sequence, and its contingency table:

C C C C C B C C C C C C B A A A A A C C C C C A A C A A B A A A B C A A A A A A
C C A A C A A A A A A A A B B C C A C A A A B B C C B A A C C C C B A A A B B A
A B C B B B C B B C B B B C C B A C A A A A B A A B A A A A C B B B B B B B B B
B B B B C A B B C C B B B B A B B B B C C B B B B B C B C A A A C C C A B B A A
B B C C C A B B B B B A B B A A C A A A B B B C B B B B B C C B B C B B B A A A
A A A A A B B B C C C C C C B B B C A A A C C A C B B B B B B B C A A B B B C C
C C C C B A B B B C C C C C C A A C C C C C C C C C C B B B B B B B B C B A C B
B A A A C B A A C C A C C C C C B B B A

A2 B2 C2
A1 54 18 17 89
B1 18 71 23 112
C1 18 23 57 98
90 112 97 299
Explanation for table: Each sequence has 300 letters total, but because the contingency table is counting the number of times that a first observed letter is followed by a second observed letter, that means are counting pairs, so there are 299 pairs. To look up the number of times that we had a pair of letters where the first was an A (i.e., A1) and the second was B (i.e., B2), we look up the corresponding cell in the contingency table; for sequence 1, ‘AB’ occurred 32 times. The yellow margin to the right sums up the total times that we saw a letter appear as the first letter in a pair (irrespective of the second), and likewise the bottom yellow margin sums up the number of times the second letter appeared irrespective of the first

Question: 3a (.5 point): Using the table for the sequence, find the marginal probabilities P(X2 = A), P(X2 = B) and P(X2 = C).

P(X2 = A) = 90/299
P(X2 = B) = 112/299
P (X2 = C) = 97/299

3b (.5 point): Suppose you followed strategy 1 and always predicted X2 = B without ever looking at X1. Of the 299 opportunities to predict X2, how often would you be right if you predict X2 = B?

P(X2 = B) = 112/299

3c (.5 point): Using the table for sequence 2, find the conditional probabilities P(X2 = A | X1 = A), P(X2 = B | X1 = B) and P(X2 = C | X1 = C).

P(X2 = A | X1 = A) = P(X2 = A and X1 = A)/ P(X1= A) = 54/89
P(X2 = B | X1 = B) = P(X2 = B and X1 = B)/ P(X1= B) = 71/112
P(X2 = C | X1 = C) = P(X2 = C and X1 = C)/ P(X1= C) = 57/98

3d (.5 point): Suppose you followed strategy 2 and predicted that X2 will be the same as X1. Use the marginal frequencies of A, B and C for sequence 2, and the conditional probabilities you just calculated (in 3c) to compute how often you would be right.

This is where I'm stuck. I have no idea what marginal probabilities they are referring to, since I already used the marginal probabilities in computing the conditional ones. Even if I manage to find that out, I have no idea whether to multiply, add or divide it with the conditional ones.

Any help would really, really be appreciated!
Thank you!
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Sun Jan 29, 2012 2:01 pm
Lava says...

Hey shloka!

If I remember correctly, you'll have to convert this to a Joint probability distribution function. The exact method I don't remember sorry.
And to get the marginal probabilities you'll use the sum of each row/column to imply MP of that variable. (Sum of all MP from all rows/columns=1)

I think this is how you do it. Forgive me if I'm wrong.
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