Solution
= Solution
After a red first draw, one red remains among six counters. The multiplication rule for <conditional probability> gives
$$
\boxed{\mathbb P(E_1\cap E_2)=\mathbb P(E_1)\mathbb P(E_2\mid E_1)=\frac27\frac16=\frac1{21}.}
$$
= Solution
After a red first draw, one red remains among six counters. The multiplication rule for <conditional probability> gives
$$
\boxed{\mathbb P(E_1\cap E_2)=\mathbb P(E_1)\mathbb P(E_2\mid E_1)=\frac27\frac16=\frac1{21}.}
$$