00:01
In this question, we can write that a production of 100 circuit boards will be here like this.
00:07
And now we can say that we are having that 6 out of 100 are defective.
00:14
So it will be here 6r and then it is here d -e -f -e -c -t -i -v -e.
00:21
And hence we can say that the value of p will be here equal to 6 divided with 100 and that is equal to 0 .06.
00:31
And hence, let a1 and a2 be the first and second board are defective with replacement.
00:38
So probability of a1 will be here equals to it will be 0 .06 and probability of a2 will also be here equals to 0 .06.
00:50
Now we can say that probability that exactly 1 is defective will be here given by, we can say exactly and then it will be here 1 is defective, that is de f -e -c, t -i -v -e, and that comes out to be equals to probability of a -1 and then it is here multiplied with 1 minus probability of a2 and then it will be here plus probability of a2 and then it is here multiplied with 1 minus probability of a1.
01:25
So solving this very expression, we can say that it will be here equals to, let us write it as 0 .06 multiplied with 1 minus 0 .06 and then here it is plus 0 .06 multiplied with it will be here 1 minus 0 .0 .06.
01:48
So the value will be here equals to 0 .1128.
01:53
So, here this same question can also be solved by binomial distribution.
01:59
So we have already found out the answer, right? an answer is this one...