Question

( (20 Marks) Wires from a manufacture are analyzed for conductivity and strength. The results from 100 wires are as follows: Strength High Conductivity (HC) Low Conductivity (LC) High (HS) 60 10 Low (LS) 18 12

Answer 1

1.The wires are given to be of 4 types. They are :

(i) High conductivity, High strength    (ii) High conductivity, Low Strength     (iii) Low conductivity, High strength   and

(iv) Low conductivity, Low strength.

(d). The first wire is any one of these 4 types. { i.e the cases are (HC1,HS1),(HC1,LS1),(LC1,HS1) and (LC1,LS1) }. Similarly for the second wire we have 4 outcomes. So the sample space has 4×4=16 elements. Thus we can describe the sample space as:

X={ (HC1,HS1,HC2,HS2),(HC1,HS1,HC2,LS2),(HC1,HS1,LC2,HS2),(HC1,HS1,LC2,LS2),(HC1,LS1,HC2,HS2),(HC1,LS1,HC2,LS2),(HC1,LS1,LC2,HS2),(HC1,LS1,LC2,LS2),(LC1,HS1,HC2,HS2),(LC1,HS1,HC2,LS2),(LC1,HS1,LC2,HS2),(LC1,HS1,LC2,LS2),(LC1,LS1,HC2,HS2),(LC1,LS1,HC2,LS2),(LC1,LS1,LC2,HS2),(LC1,LS1,LC2,LS2) }

(e) There are 16 possible outcomes in the sample space. So the probability of the given event is .

(f) For this problem, we use the concept of conditional probability.

Let A be the given event,i.e. the first wire having high conductivity and low strength. Also let B be the conditional event that the second wire has low conductivity and low strength.

We are to find ,i.e. the probability of event B such that event A is given to happen. We know that

P B A)

.----------(1)

P (BA) = P(A)

We see that      { the corresponding event being (HC1,LS1,LC2,LS2) }   and that
P(A) _ 16-4 1
because any possible outcome in event A is of the form (HC1,LS1,X,Y), where there are 2 choices for each of X and Y. Thus there are total 2×2=4 possible outcomes. Recall that the total number of outcomes is 16.

Thus plugging these values into equation (1), we have $P(B|A)=rac{rac{1}{16}}{rac{1}{4}}=rac{4}{16}=rac{1}{4}$ .

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