00:01
I'm david and i'm here to have you answer your question.
00:03
Let me bring up your question here.
00:05
Now in this question we will discuss about the binomial distribution.
00:09
Let me recall to you that if x followed by the binomial with the n and the p and then the probability of the x equal to k, it will equal to the n choose k, b power k, 1 minus b power n minus k and we have the e on the x equal to the n times b variance of the x equal to the n times b times 1 minus b now in this question we are given the exam having 10 questions it needs a multiple choice with the four principal selections for each answer and now the passing grade is called to 60 or factor suppose that the student was enabled you find the 10th history for the exam and just guesses at each question in the part i want to find the probability that he gets at least one question correct so first embecon x is a number on the correct and then the probability of the correct so this is not be the probability correct equal to 1 out of 4 and we have n here it will equal to 10 so from here x just follow the binomial with the n equal to 10 p equal to 1 over 4 or equal to 0 .25.
01:43
Now in the part i want to find the probability that he gets at least one question correct.
01:48
Means that number of the graph must be greater equal to 1.
01:52
To find this one we should tell the complement, so 1 minus probability the x equals 0.
01:57
So we have 1 minus.
01:59
We apply the formula here.
02:02
So we should get the 10, we should get the 1.
02:05
Choose 0 and then 0 .25 power 0.
02:11
1 minus 0 .25 equals 0 .75 power 10.
02:15
If we do the calculation, 1 minus that, get equal to the 0 .9, 4, 3, 6, 9.
02:30
So that will be the answer for the first one.
02:35
And then for the b, once you pass the exam, in the two parts, he must get the 60 or better.
02:44
So means that it will equal to the probability x greater equal to 6.
02:49
Now to find this one, we do it by the complement.
02:53
So it will be 1 minus probability in the x.
02:55
It's equal to 5.
02:57
So we can compute this one...