# Very Tricky Probability

John and Mary, are going to sit for an exam. The possible results that they can get are $A,B$ and $C$.

The possibility that John gets a $B$ is $0.3$. The possibility that Mary gets a $B$ is $0.4$.

The possibility that none of them gets an $A$, but at least one of them get a $B$ is $0.1$. What is the possibility that at least one of them gets a $B$, but none of them gets a $C$ ? .

So far, my approach is like:

For John get a B: $P(J_B)=0.3$
For Mary get a B: $P(M_B)=0.4$

Also, As given in the question, $[P(John_B) \cup P(Mary_B) ] ∩ P(A') = 0.1$

And I am trying to find

$[P(John_B) ∪ P(Mary_B)] ∩ P(C')$

To begin with, are my assumptions so far good/sufficient ?

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I don't see where you have made any assumptions. You have written down some formulas, but they don't make sense. If $P({\rm John\_B})=.3$ and $P({\rm Mary\_B})=.4$ then $P({\rm John\_B})\cup P({\rm Mary\_B})$ is $.3\cup.4$; how do you propose to take the union of two numbers? –  Gerry Myerson Mar 30 '12 at 11:35
As Gerry mentioned, what eactly are your assumptions? It would be better if you stated that explicitly. –  dashdart Mar 30 '12 at 12:56
hmm i see that now it makes no sense even to me.. what i wrote above... jesus –  Claus Henriksen Mar 30 '12 at 13:29
you are a lifesaver....i had the same problem....thanks Claus Henriksen credits from Creta –  user28037 Mar 31 '12 at 19:52

You are better off considering the nine individual cases, say $P_{xy}$ is probability that John gets $x$ and Mary gets $y$. You are given that $P_{BA} + P_{BB} + P_{BC} = 0.3$, $P_{AB} + P_{BB} + P_{CB} = 0.4$, $P_{BC} + P_{BB} + P_{CB} = 0.1$. so $P_{BA} + P_{BB} + P_{AB} = (0.3 + 0.4 - 0.1) = 0.6$.