# What is the probability that Jan and Jon are chosen?

Jan , Jon and $10$ other children are in a classroom. The principal of the school walks in and choose $3$ children at random. What is the probability that Jan and Jon are chosen?

My approach:

Including Jon and Jan, number of ways of selection is $1\cdot1\cdot{10\choose 1} = 10$.
Total way of selection is ${12\choose 3} = 220$

So, probability is $\frac{10}{220}= \frac{1}{22}.$

But when I solving as follows:

\begin{align*} P(\text{selection of Jon and Jan}) &= 1- P(\text{not selection of Jon and Jan}) \\ &= 1- \frac{{10\choose 3}}{{12\choose 3}}\\ &= \frac{5}{11}. \end{align*}

Which approach is correct and why alternative one is wrong?

Note: My previous post had some mistakes, so I deleted that

I will call $A$, $B$ the events selecting Jon, Jan respectively. You did not take the complement correctly \begin{align*} P(A\cap B) &= 1-\color{red}{P(\bar A \cup\bar B)}\tag 1 \\ &= 1-[P(\bar A)+P(\bar B)-P(\bar A\bar B)]\tag 2 \\ &=1-\left[\frac{\binom{11}{3}}{\binom{12}{3}}+\frac{\binom{11}{3}}{\binom{12}{3}}-\frac{\binom{10}{3}}{\binom{12}{3}}\right]\\ &=\frac{1}{22} \end{align*} where $(1)$ is true by DeMorgan's law, and $(2)$ is true by inclusion-exclusion. As you can see the two methods give the same value.

Your first approach gives the probability that Jon AND Jan are selected, the second gives the probability that Jon OR Jan are selected.

When you computed the probability of not selecting Jon AND Jan, you didn't include the situations where Jon was selected but not Jan, and vice versa.

Lets say the even Jon is selected is A, Jan is selected B.

Then $P(A \wedge B)=1-P(Not (A \wedge B))= 1- P(Not A \vee Not B)$

$P(Not A \vee Not B)= P(Not A)+ P(Not B)- P(Not A \wedge Not B)$

Here $\wedge$ means and, and $\vee$ means or. Think of a Venn diagram, if we look at the union of two circles, the total area is equal to the sum of the circles minus the intersection, because we counted that part twice.

$P(Not A)=1-P(A)=1-(1*\binom{11}{2})/220)=3/4$

or

$P(Not A)=\binom{11}{3}/220=3/4$

$P(Not B)=3/4$

$P(Not A \wedge Not B)= \binom{10}{3}/220=6/11$

Now $3/4+3/4-6/11=21/22$ So $1-21/22=1/22$