Hi Guys

Please see attached the joint probability distribution.

X = amount of winnings for B (0 represents a tie)

Y= Whether A Studied (represented by 0) or not studied (represented by 1)

A bets B that A will outscore B in an exam. The joint probability distribution is the attached picture.

What is the probability distribution going to look like if B KNOWS that A DID NOT study.

I can't seem to make headway on this one. Conceptually, I know that if A did not study, B loosing (X = -10) will be zero, however, A and B may tie (as opposed to currently zero) and that B winning will have a higher chance than 0.45.

The fields in this distribution table that were given were: 0.75, 0.45,0.3 and 0.18. I calculated the rest.

Input is appreciated.

On another try, here is what I thought:

Since B loosing will have zero probability if A did not study, and B winning 10 has probability 0.45 if A did not study, means the remaining probability 0.55 has to be A & B coming to a tie.

Please see attached the joint probability distribution.

X = amount of winnings for B (0 represents a tie)

Y= Whether A Studied (represented by 0) or not studied (represented by 1)

A bets B that A will outscore B in an exam. The joint probability distribution is the attached picture.

What is the probability distribution going to look like if B KNOWS that A DID NOT study.

I can't seem to make headway on this one. Conceptually, I know that if A did not study, B loosing (X = -10) will be zero, however, A and B may tie (as opposed to currently zero) and that B winning will have a higher chance than 0.45.

The fields in this distribution table that were given were: 0.75, 0.45,0.3 and 0.18. I calculated the rest.

Input is appreciated.

On another try, here is what I thought:

Since B loosing will have zero probability if A did not study, and B winning 10 has probability 0.45 if A did not study, means the remaining probability 0.55 has to be A & B coming to a tie.

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