The police told Mr.K that 5/8 students in his class own the gun. That is a hot news for Mr.K .Immediately, he picks 10 of his students and checks each of them. a/What is the probability that he will find out no one own the gun? b/The students only use their toys if Mr.K finds out half or more than half of 10 students he picked own the gun. What is the probability that the students, who are not own the gun cannot do their homework that night? (The probability that the students own the gun use their gun) By TN
The police told Mr.K
I have a HOT NEWS for you I win 649 ???
Oh! Really? NO! 5/8 students in your class own the gun. !!!
You are number 1, you are number 2 …you are number 10..All of you come here … Immediately, he picks 10 of his students and checks each of them. Huh???... Me ?? He’s so cute!!! ??
a/ What is the probability that he will find out no one own the gun???
Don’t own the gun
There are exactly 2 possible out come Binomial Distribution
Own the gun
We know 2 cases when Mr.K check each student in his 10 students
Data is discrete ( number of student own the gun in 10 students )
When a Binomial Experiment is conduct many, many time, its histogram will approach the shape of the normal curve.
Recall Binomial Distribution
We have so many ways to solve this problem, let’s star at the basic way !! First, we call : O : own the gun N : don’t own the gun Second, draw the tree diagram : N O N O N O N O N O N O N O … … … … And so on…
N O N O N O N O N O N O N O … … … … Then we put probabilities in the tree diagram … 5/8 5/8 5/8 3/8 3/8 3/8 3/8 3/8 3/8 3/8 5/8 5/8 5/8 5/8 No one own the gun, it’s that is the last branch of the tree Therefore, the answer is P(NNNNNNNNNN) = (3/8)^10 = 5.499 * 10^-5 (5.499 * 10^-3 %)
Another way… Permutation of non-Distinguishable object We also assume: O : own the gun N : don’t own the gun No one own the gun => We will have 10 letters 10 : NNNNNNNNNN How many ways to arrange that contain 10 letters “N” and can not repeat ?
we will have the equation 10! / 10! = 1 Total letters Total letter “N” be repeat => Probability = 1 * [ (5/8) ^ 0 ]*[ (3/8) ^ 10 ] = 5.499*10^-5 (5.499*10^-3 %) Probability of the students own the guns Probability of the students don’t own the guns 0 student own the gun 10 students don’t own the guns
Another way, it is seem easy than the First and Second ways is use calculator (TI-83) This is the formula Binompdf(X,Y,Z) X: number of trials Y: probability success Z: specific outcome Binompdf(X,Y,Z) [2 nd ]=>[VARS]=> move UP [^] or DOWN [v] => binompdf(X,Y,Z) So, set up the calculator screen like this ….
binompdf(10,5/8,0) The answer is 5.499*10^-5 (5.499*10^-3 %)
Which mean we need to find the probability that the students own the gun use their gun b/The students only use their toys if Mr.K finds out half or more than half of 10 students he picked own the gun. What is the probability that the students, who are not own the gun cannot do their homework that night?
We can use “ tree diagram ” or “ Permutation of non-Distinguishable object ” or “ binompdf ” to solve this problem by find each probability for 1 student own the gun to 5 students own the guns, and add them together. But these are not the good choices ^_^ The best choice to solve this problem is : binom cdf ..At first
THE DIFFERENCE BETWEEN BINOMPDF AND BINOMCDF Binompdf(X,Y,Z) Binomcdf(X,Y,z) It will give exactly the probability of Z It will add up the probability of 0 to probability of z AAA BAA BBA BBB AAA BAA BBA BBB If Z = 3 If z= 3 It will give the probability of this outcome It will give the total probability from AAA to BBA . Note : Binomcdf will add the probability from left to right
Add from here To z EX :
First, we need to know the probability of at most 4 (include) students in 10 students own the guns Binomcdf(10,5/8,4) =0.1275 (12.75%) Probability of at most 4 students own the gun Add the probability up from non of the student to 4 students who own the gun This is also the “safe probability”
So the probability that 5 or more than 5 students own the gun is : 1- 0.1275 =0.87249(87.25%) DANGEROUS !!