Bayes Theorem where the number of tosses depends on tosses There are two fair coins (i.e. Heads and Tails are equally likely for tosses of both). Coin
$1$ is tossed $3$ times. Let $X$ be the number of Heads that occur. After this, Coin $2$ is
tossed $X$ times. Let $Y$ be the number of Heads we get with Coin $2$. The probability
$Pr(X ≥ 2|Y = 1)$ equals ?
I tried to solve it using Bayes' theorem such that
$Pr(X\geq 2|Y=1)={Pr(X\geq 2)\cdot[Pr(Y=1|X=2)+Pr(Y=1|X=3)]\over Pr(X>=2)\cdot[Pr(Y=1|X=2)+Pr(Y=1|X=3)]+Pr(X<2)\cdot[Pr(Y=1|X=0)+Pr(Y=1|X=1)]} $
and i got the answer $7/11$ which seems to be wrong. Can anyone help me understand why this method fails to work?
 A: $$P(X\geq 2 | Y=1) = \frac{P(X \geq 2 \text{ and } Y=1)}{P(Y=1)}$$
We can solve for the numerator and the denominator separately.
$X$ can be $0,1,2,$ or $3$.
$Y$ can equal $1$ in $3$ different ways, because if $X=0$ then $Y$ cannot be $1$.  Each of those is a product, so the denominator is a sum of three terms.
The numerator is easier.  We can split that up into
$$P(X \geq 2 \text{ and } Y = 1) = P(Y = 1 | X \geq 2) \cdot P(X \geq 2)$$
The numerator is two cases, if $X=2$ and if $X = 3$, so the numerator will be a sum of two products:
$$P(Y=1|X=2)\cdot P(X=2) + P(Y=1|X=3)\cdot P(X=3)$$
From there you can put the answer together.  It's hard to read your equation, but I think your problem is that you were trying to factor out a term where there is no common factor.  You have to compute each specific product separately, then add.


I tried to solve it using Bayes' theorem such that
$Pr(X\geq 2|Y=1)={Pr(X\geq 2)\cdot[Pr(Y=1|X=2)+Pr(Y=1|X=3)]\over Pr(X>=2)\cdot[Pr(Y=1|X=2)+Pr(Y=1|X=3)]+Pr(X<2)\cdot[Pr(Y=1|X=0)+Pr(Y=1|X=1)]} $

Problem: $\Pr(Y=1\mid X\geq 2)\neq \Pr(Y=1\mid X=2)+\Pr(Y=1\mid X=3)$ as the Additive Law applies to unions of disjoint events, not conditions.
Solution: Deal with the union before the conditioning.
In summary:
$\begin{align}\Pr(X\geq 2\mid Y=1)&=\dfrac{\Pr(X=2,Y=1)+\Pr(X=3,Y=1)}{\Pr(X=1,Y=1)+\Pr(X=2,Y=1)+\Pr(X=3,Y=1)}\\[1ex]&=\tfrac{\Pr(X=2)\Pr(Y=1\mid X=2)+\Pr(X=3)\Pr(Y=1\mid X=3)}{\Pr(X=1)\Pr(Y=1\mid X=1)+\Pr(X=2)\Pr(Y=1\mid X=2)+\Pr(X=3)\Pr(Y=1\mid X=3)}\\[1ex]&=\dfrac{5}{9}\end{align}$
