00:01
In this question we are given that are two typists and your article is typed.
00:06
Now your article is equally likely to be typed by either typist.
00:11
I'm going to let t1 denote the event the first type is type your article and t2 denote the event the second type is type your article.
00:20
Probability of t1, that is the first type is type your article, would be 0 .5 since it's equally likely to be typed by either type is.
00:31
And probability of t2, that is articles type by the second type is, would be also 0 .5.
00:40
Now for the first type is, the average number of errors made per article is 2 .1.
00:46
For second type is the average number errors made per article is 3 .3.
00:51
I'm going to let e1 denote the number of errors in a given article type by the first type is.
00:58
And e2 denote the number errors in a given article type by the second type piece.
01:05
So e1 follows the poison distribution, the mean is 2 .1, and e2 follows the poison distribution.
01:17
The mean is 3 .3.
01:19
For e1, probability of e1 equals to r.
01:25
There is there are r number errors in the article typed by the first typist and that would be 2 .1 to power r over r factorial times e to power minus 2 .1.
01:41
Probability of e2 equals to k where k is the number of errors typed by the second typist and that would be 3 .3 to power k over k factorial times e to power minus 3 .3.
02:09
We want to find probability your article will have no errors...