00:01
To solve this question, first we need to know that if we have two switches, 1 and 2 that operate independently, then the probability of closing, it is p1, p2.
00:30
Here's the case that they are in series connection.
00:35
If the two switches are in parallel connection like this, then the probability of closing, it is the probability that one is closing or two is closing.
01:06
So it is p1 plus p2 minus p1 p2 by the inclusion exclusion principle.
01:34
Now we are going to deal with part a of our question.
01:38
So the upper half, this is a series connection.
01:44
So we have p1, p2, the probability of closing.
01:54
And for the lower part, we have p3, p4, for the probability of closing.
02:08
But the upper half and the lower half, they are in parallel connection, so we need to sum them and the sub -chapter.
02:17
Subtract the product.
02:25
So this is the left system of this circuit.
02:41
And the right hand side we have a 5.
02:44
It is in a series connection with the left hand side part, so we just multiply p5.
02:54
So this should be the answer of part a.
03:00
So what we need to do is just plug in the value of p1, p2, 3, b.
03:05
For p5 so there are 0 .6 times 0 .7 plus 0 .5 times 0 .9 minus their product and then times p5 which is 0 .1.
03:36
And the answer of this is 0 .0681.
03:46
Now we go on to part b.
03:49
In part b, in part we need to consider separate cases.
03:55
The first case is 3 is closed, and the second case 2 is 3 is open...