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Suppose $X,Y,Z$ are independent binary random variables taking values in ${0,1}$. Let $P(X=0)=p$, $P(Y=0)=q$, and $P(Z=0)=r$. Which expression is equal to $P(X+Y+Z\geq 2\mid X+Y\geq 1)$?

  1. $\frac{(1-p)(1-q)+pq(1-r)}{1-pq}$
     
  2. $\frac{(1-p)(1-q)+(1-r)[p(1-q)+q(1-p)]}{1-pq}$
     
  3. $\frac{(1-p)(1-q)r+(1-r)[p(1-q)+q(1-p)]}{1-pq}$
     
  4. $\frac{1-pqr}{1-pq}$

1 Answer

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We need $P(X+Y+Z\geq 2\mid X+Y\geq 1)$.

Using conditional probability, this is $\frac{P(X+Y+Z\geq 2,\ X+Y\geq 1)}{P(X+Y\geq 1)}$.

Since $X+Y+Z\geq 2$ already forces $X+Y\geq 1$ except in the case $X=0,Y=0,Z=2$, which is impossible because $Z\in{0,1}$, the numerator is just $P(X+Y+Z\geq 2)$.

Now $P(X+Y\geq 1)=1-P(X=0,Y=0)=1-pq$.

The event $X+Y+Z\geq 2$ happens in these cases: $X=1,Y=1,Z=0$, $X=1,Y=0,Z=1$, $X=0,Y=1,Z=1$, or $X=1,Y=1,Z=1$.

So the numerator is $(1-p)(1-q)r+(1-p)q(1-r)+p(1-q)(1-r)+(1-p)(1-q)(1-r)$.

Combine the first and last terms: $(1-p)(1-q)r+(1-p)(1-q)(1-r)=(1-p)(1-q)$.

So the numerator becomes $(1-p)(1-q)+(1-r)[p(1-q)+q(1-p)]$.

$\boxed{\therefore P(X+Y+Z\geq 2\mid X+Y\geq 1)=\frac{(1-p)(1-q)+(1-r)[p(1-q)+q(1-p)]}{1-pq}}$

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