What is the probability that you a fair die lands on 1 or 2, given that you rolled a number strictly less than 4?
While the solution give as 2/3. but I can't get the understanding right.
here is my thinking: P( 1or2 | l4 ) = P( l4 | 1or2 )P(1or2)/P(l4) but when i plug in P(1or2) =1/3 and P(l4)=2/3 and P(l4|1or2) = 1.( pretty sure this one doesn't look right) . Can get right. as 2/3 as the given solution. Anyone please give me a hand. Thank you in advance.
Your error is in saying P(l4) is $\frac{2}{3}$. Notice, that of the six equally likely sides to a standard die, exactly three of those sides are strictly less than $4$, and so the probability of having rolled a number strictly less than for is $\frac{3}{6}=\frac{1}{2}$, not $\frac{4}{6}=\frac{2}{3}$. Having rolled a $4$ does not count as having rolled a number strictly less than $4$.
In making this correction, we get as an answer $\dfrac{~\frac{1}{3}~}{\frac{1}{2}}=\dfrac{2}{3}$ as expected.