WEBVTT

1
00:00:00.000 --> 00:00:08.417
Here is the problem: An event has 6 favorable outcomes out of 36 equally likely outcomes. What is its probability?

2
00:00:08.867 --> 00:00:11.491
Single event (favorable divided by total).

3
00:00:12.291 --> 00:00:16.204
Divide favorable outcomes by total equally likely outcomes.

4
00:00:16.654 --> 00:00:18.790
Complement: the event does not happen.

5
00:00:19.240 --> 00:00:20.355
Odds in favor.

6
00:00:21.155 --> 00:00:30.745
So the answer is P(A) equals 1 over 6 approximately 0.16666667 (16.666667 percent).

7
00:00:31.195 --> 00:00:45.928
The event occurs in 6 of 36 equally likely outcomes, so its probability is 16.666667 percent; it fails with probability 83.333333 percent.