1996 United States presidential election in Washington (state)|
|
Turnout | 74.52%
8.08%[1] |
---|
|
![](//upload.wikimedia.org/wikipedia/commons/thumb/4/47/Washington_Presidential_Election_Results_1996.svg/350px-Washington_Presidential_Election_Results_1996.svg.png) County Results
Clinton
40-50%
50-60%
|
Dole
40-50%
50-60%
| |
|
The 1996 United States presidential election in Washington took place on November 5, 1996, as part of the 1996 United States presidential election. Voters chose 11 representatives, or electors to the Electoral College, who voted for president and vice president.
The State of Washington was won by President Bill Clinton (D–AR) over Senator Bob Dole (R–KS), with Clinton winning 49.84% to 37.30% for a margin of 12.54%. Billionaire businessman Ross Perot (Reform–TX) finished in third, with 8.92% of the popular vote.[2] As of the 2020 presidential election[update], this is the last election in which Spokane County, Kittitas County, Pend Oreille County, Ferry County, and Asotin County voted for a Democratic presidential candidate.[3]