00:01
Okay, this is the question that we have given.
00:03
Suppose that a box contains eight cameras, eight cameras, and that four of them are defective.
00:13
A sample of two cameras is selected at random, and we have to define the random variable x as the number of defective cameras in the sample.
00:22
Write the probability distribution for x.
00:26
Okay, now we have to define the random variable x as the number of defective cameras, and also we should have to write the probability distribution for x now let us see how to do this problem total cameras in the box total cameras equal to eight in the box in that four of them are defectives therefore defective cameras defective cameras i'm defining it as d equal to four since four are defective cameras the the remaining camera should be non -defective therefore non -defective cameras non -defective cameras i'm defining it as n d non -defective is equal to four sorry non -defective cameras equal to total cameras minus defective cameras 8 minus 4 that is nothing but 4 okay now we are asked to find a random variable define the random variable x as the number of defective cameras in this sample in the sample we have to select two cameras so now let us try to define the random variable x i am writing with tender pen let x be the number of defective cameras number of defective cameras therefore what are the values x can take since you have to select only two cameras out of the two cameras we can have zero defectives we can have only one defect two we can have maximum of two defective cameras more than that it is not possible to define the random variable x so here the x is random variable and it can take values of 0 1 2 since x is number of defective cameras in the sample we have to select two cameras so out of two cameras the possibility is none of them is defective that means no cameras defective and only one camera as defective and maximum of two cameras can be defective other than that there is no possible values that x can take so this is the random variable x now let us try to define the probability p of x for x so by finding the probability we complete the probability distribution the second part of our question now let us see how we can do that now as you know that the basic definition for probabilities so i will write with tender color so it will be easy for you.
03:28
Okay.
03:30
As you know that probability, the basic definition for probability, probability p equal to number of favorable outcomes by total outcomes.
03:59
So if you do any experiment, the probability can be defined as the ratio of favorable outcomes to the total outcomes.
04:08
It is the basic definition of probability.
04:10
Now with the help of this, let us try to find the probability distribution.
04:15
Okay.
04:17
Okay.
04:18
So i am drawing a table.
04:21
This is the probability distribution table for the random variable x.
04:30
So these are the values x can take and p of x is the probability.
04:35
Now let us see how we can determine the probability.
04:39
Okay.
04:39
Suppose first one when x equal to 0.
04:49
What is x equal to 0? x equal to 0 implies there are no defective cameras.
04:56
That is two cameras are non -defective.
05:06
X equal to 0 tells us that two cameras in the sample are not defective.
05:11
So the probability for selecting such two cameras is since two cameras are non -defective, we can select the two cameras from this 4 only from this 4 so the favorable outcomes is 4 c2 because we can select two cameras from non -defective it is 4 and the total outcomes is 8c2 because from 8 we can select 2 so that has a total outcomes but we don't want that the favorable outcomes is that non -defective cameras so if you do this we get it as 3 by 14 so we can write here for x equal to 0 random variable the probability is 3 by 14 now let us see the other one the other one is x equal to 1 what is x equal to 1 represents it represents one camera is defective one camera is defective and one camera is non -defective because we have to select two cameras...