Question

A defibrillator manufacturer uses three different materials (Silver, Titanium, and Copper) for the electrodes in its devices. Each electrode is made solely from a single material. The electrodes are designed as an adhesive pad, which must carry strong pulses of electricity without deteriorating and remain functional after 10 years. Events are defined as follows: .Ag: the electrode is made of Silver Ti: the electrode is made of Titanium . Cu: the electrode is made of Copper . X: the electrode is functional after 10 years The probabilities that an electrode is made out of the different materials are as follows:
Ag: the electrode is made of Silver
Ti: the electrode is made of Titanium
Cu: the electrode is made of Copper
X: the electrode is functional after 10 years

The probabilities that an electrode is made out of the different materials are as follows:
P(Ag) 0.2 ; P(Ti) 0.7; P(Cu) 0.1
and the conditional probabilities of functionality after 10 years are as follows:

P(XlAg)=0.9, P(XlTi)= 0.8, P(X l Cu)=0.6


i. Using words, explain what P(Ag|X) represents. Then, compute P(Ag|X).
ii. Using words, explain what P(Ag,IX) represents. Then, compute P(Ag'IX).
iii. Using words, explain what P(X) represents. Then, compute P(X).

Answer 1

Answer:

i. P(Ag|X) is the probability of the electrode being made of silver, given that it is still functional after 10 years. P(Ag|X)= 0.225

ii.  P(Ag'IX) is the probability that the electrodes arent made of cooper given that they still work after 10 years.  P(Ag'IX)= 0.78

iii. P(X) is the probability that the electrodes are still functional after 10 years. P(X)= 0.8

Step-by-step explanation:

Hello!

All of them are conditional probabilities.

i. Using words, explain what P(Ag|X) represents. Then, compute P(Ag|X).

P(Ag|X) is the probability of the electrode being made of silver, given that it is still functional after 10 years, i.e. both events are dependent, the occurrence of "X" modifies the probability of "Ag" to happen.

$P(Ag|X) = \frac{P(AgnX)}{P(X)}$

There are two values that we don't know, one is the probability of the intersection between Ag and X, symbolized P(Ag∩X) and the other one is the probability of X, symbolized P(X)

To calculate the probability of the intersection we will use the following known values:

P(Ag) 0.2 and P(XlAg)=0.9

Now if $P(XlAg)= \frac{P(XnAg)}{P(Ag)}$

Then $P(XnAg)= P(X/Ag)*P(Ag)$

$P(XnAg)= 0.9*0.2= 0.18$

To know the probability of "X" you have to add all the events where it occurs. To see which are those events, arrange all possible events of this experiment in a contingency table:

         Ag              Ti           Cu          Total

X ; P(Ag∩X) ; P(Ti∩X) ; P(Cu∩X) ; P(Ag∩X) + P(Ti∩X) + P(Cu∩X)= P(X)

X' ;P(Ag∩X') ; P(Ti∩X') ; P(Cu∩X') ; P(Ag∩X') + P(Ti∩X') + P(Cu∩X')= P(X')

    P(Ag∩X)  ; P(Ti∩X)  ; P(Cu∩X)

T.         +       ;       +       ;        +

    P(Ag∩X') ; P(Ti∩X')  ; P(Cu∩X')

          =        ;       =       ;         =

         P(Ag)  ;    P(Ti)     ;     P(Cu)

Using the probabilities of  P(Ti) 0.7 and  P(XlTi)= 0.8 you'll obtain P(Ti∩X):

$P(TinX)= P(X/Ti)*P(Ti)= 0.8*0.7= 0.56$

And using P(Cu) 0.1 and P(X l Cu)=0.6 you'll obtain P(Cu∩X)

$P(CunX)= P(X/Cu)*P(Cu)= 0.1*0.6= 0.06$

Then the probability of X is:

$P(X)= P(AgnX)+P(TinX)+P(CunX)= 0.18+0.56+0.06= 0.8$

Now you can calculate the asked probability:

$P(Ag|X) = \frac{P(AgnX)}{P(X)}= \frac{0.18}{0.8}= 0.225$

ii. Using words, explain what P(Ag'IX) represents. Then, compute P(Ag'IX).

Ig "Ag" means that the electrodes were made of silver then "Ag'" is its complimentary event and means that the electrodes aren't made of silver. (they are either made of Titanium or Copper).

P(Ag'IX) is the probability that the electrodes aren't made of cooper given that they lasted for 10 years.

P(Ag')= P(Ti∪Cu)= P(Ti)+P(Cu)

And if you watch the table I've made you can calculate the intersection of Ag' and X as:

P(Ag'∩X)= P(Ti∩X) + P(Cu∩X) = 0.56+0.06= 0.62

$P(Ag'IX)= \frac{P(Ag'nX)}{P(X)}= \frac{0.62}{0.8}= 0.78$

iii. Using words, explain what P(X) represents. Then, compute P(X).

P(X) is the probability that the electrodes are still functional after 10 years.

I hope it helps!