00:01
In the first part of this question we are given that the total number of capacitors is 100.
00:14
The number of capacitors within the tolerance values is 73.
00:28
The number of capacitors below the required tolerance level is 17 and the remaining capacitors are above the required tolerance level.
00:47
So the remaining capacitors will be calculated as 100 minus 73 plus 17, that is 10.
01:02
In the first part, we have been asked to find the probability of selecting two capacitors, one after the other, and both should be within the required tolerance level along with replacement.
01:17
So in this case, in the case of replacement, the probability will be calculated as here the probability p will be within the tolerance value we have 73.
01:33
So 73 divided by 100 multiplied by 73 divided by 73 divided by 100.
01:48
This comes out to be 5329 divided by 1 4 times 0.
02:03
Thus the required probability is 0 .5329.
02:18
Now in the second part we have to find the probability of selecting two capacitors.
02:24
When the first one is drawn, and it is below the required tolerance level and the second one is above the required tolerance level and here there is no replacement.
02:44
So let's calculate the probability in this case.
02:49
So the first capacitor is below the required tolerance level.
02:56
So the capacitors that are below the tolerance level there are only 17.
03:00
So 17 divided by 100.
03:03
Multiplied by second one is above required tolerance level that means 10 divided by 99 99 because there is no replacement so we get 170 divided by 9900 which is 17 divided by 990 which comes out to be 0 .0172.
03:44
Thus, the required probability is 0 .017.
03:57
Now in the second situation we are given that the total number of lamps is 45.
04:07
The number of faulty lamps is 10 and remaining are satisfactory lamps...