<s>[INST] First coin is tossed 3 times. $X$=heads after 3 tosses.Second coin is tossed $X$ times and$Y$represents total heads.Find $P(X>=2|Y =1) $ There are 2 fair coins. First coin is tossed 3 times. $X$ represents number of heads in the 3 tosses.After this second coin is tossed $X$ number of times, where $Y$ represents number of heads.Find $P(X>=2|Y =1) $
Here, by statistical reasoning it can be concluded that the answer is $(5/9)$, but it appears to be $(11/18)$. 
My attempt: 
P(X greater than or equal to 2 given Y=1)=1- P(X is 0 or 1 given Y=1)
Which is, 
= 1- P(X is one given Y is one)
= $1- (4/9)$
Which is indeed (5/9). 
 [/INST] Can you tell me how does it appear to be $\frac{11}{18}$? Is it by simulation or something else? I'm giving the mathematical argument below. Your conclusion based on statistical reasoning is absolutely fine! There seems to be some holes in whatever procedure that gives you $\frac{11}{18}$ as the answer.
First you need the joint distribution of $(X,Y)$. Note that
\begin{align}
P(X=x, Y=y)&=P(Y=y|X=x)P(X=x)\\
&=\binom{x}{y}\bigg(\frac{1}{2}\bigg)^x\cdot\binom{3}{x}\bigg(\frac{1}{2}\bigg)^3\\
&=\binom{x}{y}\binom{3}{x}\bigg(\frac{1}{2}\bigg)^{x+3}
\end{align}
Armed with the joint distribution, now we can calculate
\begin{align}
&\frac{P(X \geq 2, Y=1)}{P(Y=1)}\\
&=\frac{P(X=2,Y=1)+P(X=3,Y=1)}{P(X=0,Y=1)+P(X=1,Y=1)+P(X=2,Y=1)+P(X=3,Y=1)}\\
&=\frac{\binom{2}{1}\binom{3}{2}(\frac{1}{2})^5+\binom{3}{1}\binom{3}{3}(\frac{1}{2})^6}{0+\binom{1}{1}\binom{3}{1}(\frac{1}{2})^4+\binom{2}{1}\binom{3}{2}(\frac{1}{2})^5+\binom{3}{1}\binom{3}{3}(\frac{1}{2})^6}\\
&=\frac{5}{9}
\end{align}
</s>