# Let $P(X=1)=0.6$ and $P(X=1,Y>=5)=0.2$. Find $P(X=1,Y<5)$.

I'm thinking with the given information $P(Y\geq5) = P(X=1,Y\geq5)/P(X=1)$, so $P(Y\geq5)=0.33$. Therefore, $P(Y<5) = 1-P(Y\geq5)$, so $P(Y<5)=0.67$. Which means $P(X=1,Y<5) = P(Y<5)P(X=1) = 0.402$.

Is this correct? If not, how is it done?

They are not independent.

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Perhaps you might like to ask a single question and to engage with those that answer it before asking a series of questions. I'm sure that you would be better equipped to answer your second question if you spent time engaged with you first question. –  Fly by Night Oct 9 '12 at 17:38

Let $A$ denote the set of all $x, y$ values such that $X = 1$. Let $B$ denote the set of all $x, y$ values such that $X = 1, Y \geq 5$. Let $C$ denote the set of all $x, y$ values such that $X = 1, Y < 5$. Then, we have $A = B \cup C$ and $B$ and $C$ are disjoint, i.e., they do not overlap. Therefore,
$$P(A) = P(B) + P(C)$$
This is a general principle you should make sure you remember and understand. Spelling out what $A, B, C$ mean, we have:
$$P(X = 1) = P(X=1, Y \geq 5) + P(X = 1, Y < 5)$$
Independence has nothing to do with this problem, and finding $P(Y \geq 5)$ is not applicable and not possible with the given information. The only pairs of $x, y$ we have any information about are those where $x = 1$, so we can not possibly say anything about $Y$ in general.