Probability

G

Genevie

New member
Joined
Apr 29, 2020
Messages
6
A random sample of 6 items is taken from a large consignment and tested in two independent stages. The probability that an article will pass either stage is q. All six items are first tested at stage 1, and provided 5 or more pass, those which pass are retested at stage 2. The consignment is accepted if there is no more than one failure at each stage. Find expressions in terms of q for:

a) The probability that stage 2 of the test will be required.

b) The number of items expected to undergo stage 2.

c) The probability P(q) of accepting the consignment.

(d) Show that dP d q = 0 when q = 1 and find P(q = 0.9) and P(q = 0.8). Comment on your results.



a) P(stage2) = P(x>=5) = P(x=5) + P(x=6)
But how do I right in terms of q?

b) (a) E(x>=5) = E(5)+E(6)

c) P(X=x) = q^x * (1-q)^1-x

d) do I then replace those q’s in the formula above





Please help, I’m not sure of my answers
 
a)
The formula for P(x=r) is nCr*p^r*(1-p)^(n-r)

P(x=5)=6C5*q^5*(1-q)^1 and P(x=6)=6C6*q^6*(1-q)^0
Expanding and simplifying, you should get P(x>=5)=-q^7-5q^6+6q^5
 
Last edited:
Correction
a)
The formula for P(x=r) is nCr*p^r*(1-p)^(n-r)

P(x=5)=6C5*q^5*(1-q)^1 and P(x=6)=6C6*q^6*(1-q)^0
Expanding and simplifying, you should get P(x>=5)=6q^5-5q^6
 
c) The probability P(q) of accepting the consignment.
P1(x=5)*P2(x=5)+(P1(x=6)*P2(x>=5)
(6q^5-6q^6)*q^5+q^6*(6q^5-5q^6 )=6q^10-5q^12

d) Show that dP d q = 0 when q = 1

From a) we have that P(x>=5)=6q^5-5q^6. The derivative of that is 30q^4-30q^5
When q=1, 30*1-30*1=0

From c, the probability of passing is 6q^10-5q^12. When q=0.9:

6*0.9^10-5*0.9^12= 0.6799

When q=0.8

6*0.8^10-5*0.8^12=0.3006
 
a)
The formula for P(X=x) is actually nCx*p^x*(1-p)^(n-x)

You missed the nCr. Then you can sub as you mentioned p=q, n for the total of elements and x for the wished outcome
 
b)
You can apply directly the formula of the expected value on a Binomial (n,p) as E=n*p
 
You must log in or register to reply here.
Top