DFA for: even number of 0s and odd number of 1s over $\{0,1\}$

Idea: track two parities, the parity of the number of 0s and the parity of the number of 1s seen so far. That gives $2 \times 2 = 4$ states.

State	#0s	#1s
$A$ (start)	even	even
$B$	odd	even
$C$ (final)	even	odd
$D$	odd	odd

Reading a 0 flips the parity of 0s; reading a 1 flips the parity of 1s. Accepting state: $C$ (even 0s, odd 1s).

Transition table

State	on 0	on 1
$\to A$	$B$	$C$
$B$	$A$	$D$
$*C$	$D$	$A$
$D$	$C$	$B$

Transition diagram

Formally $M = (\{A,B,C,D\}, \{0,1\}, \delta, A, \{C\})$.

Verification

Input	Path	Final state	0s / 1s	Result
1	$A \to C$	$C$	0 / 1	Accepted
001	$A \to B \to A \to C$	$C$	2 / 1	Accepted
10011	$A \to C \to D \to C \to A \to C$	$C$	2 / 3	Accepted
01	$A \to B \to D$	$D$	1 / 1	Rejected
0110	$A \to B \to D \to B \to A$	$A$	2 / 2	Rejected
11	$A \to C \to A$	$A$	0 / 2	Rejected
$\varepsilon$	$A$	$A$	0 / 0	Rejected

Every accepted string has an even count of 0s and an odd count of 1s, and every rejected one violates at least one condition, so the DFA is correct.