Calculated probability is different from given answer This problem is from Probability and Statistics - The Science of Uncertainty, Second Edition. 
Suppose we choose a positive integer at random, according to some unknown
probability distribution. Suppose we know that P({1, 2, 3, 4, 5}) = 0.3, that P({4, 5, 6})
= 0.4, and that P({1}) = 0.1. What are the largest and smallest possible values of
P({2})?
The answer given to me states that P({2}) could be between 0 and 0.2, however my solution says that P({2}) is between 0 and 0.162. I got this by:


*

*Getting P({1,2,3}) by multiplying P({1, 2, 3, 4, 5}) with P({4, 5, 6}^c), which got me 0.18.

*Then I got P({2,3}) by multiplying P({1,2,3}) with P({1}^c), which got me 0.162.

*Therefore, since P({2,3}) was 0.162, I said P({2}) was between 0 and 0.162.
My question is where in my solution I went wrong and how to fix it, or whether the solution given to me was wrong.
Thank you.
 A: "Then I got $P(\{2,3\})$ by multiplying $P(\{1,2,3\})$ with $P(\{1\}^c)$"
Who said you could multiply these?
Note:  $Pr(A\cap B) = Pr(A)\times Pr(B)$ is true if and only if $A$ and $B$ are independent events.  When they are not independent, the above is actually an inequality, not an equality.

A corrected approach:
We know that $P(\{1,2,3,4,5\})=0.3$ and that $P(\{1\})=0.1$, so $P(\{2,3,4,5\})=0.3-0.1=0.2$
We know that $\{2\}\subseteq \{2,3,4,5\}$ so $P(\{2\})\leq P(\{2,3,4,5\})=0.2$
Now, the question is if this bound is strict... it will be.  Consider the case where $P(\{1\})=0.1, P(\{2\})=0.2, P(\{6\})=0.4$ and $P(\{7\})=0.3$ with $P(\{x\})=0$ for all other values of $x$.  It is clear to see that this satisfies all of the necessary conditions.
As for the lower bound, every probability naturally has a lower bound of $0$.  The question is if it is strict here.  Indeed, using the same example as above but replacing the role of $2$ and $3$, setting instead $P(\{2\})=0$ and $P(\{3\})=0.2$, we again satisfy all of the desired conditions.
