Solving Basic Probability Homework: Can't Understand Solution

[Clara Chung]

Homework Statement

Homework Equations

The Attempt at a Solution

I can't understand the solution.
Shouldn't the probability be
0.99^2 x (0.96^2 + 0.96(0.04) + 0.04^2) + (0.01 (0.99) +0.01^2) x (0.96^2).
Please someone explain the solution to me

 

[BvU]

Hi,

So what are the relevant equations for (a) and (b) ?

can't understand the solution

Which one? There are two solutions !

 

[Clara Chung]

Hi,

So what are the relevant equations for (a) and (b) ?
Which one? There are two solutions !

I don't understand part a. I have no idea.
I think the relevant equations for (a) and (b) are simply basic probability rules like P(A) X P(B) = P(A and B)
thanks

 

[Stephen Tashi]

I can't understand the solution.
Shouldn't the probability be
0.99^2 x (0.96^2 + 0.96(0.04) + 0.04^2) + (0.01 (0.99) +0.01^2) x (0.96^2).
Please someone explain the solution to me

You didn't explain your own solution. I can't tell what you are thinking.

The text is computing:
Pr( lot accepted) =
Pr(lot is from B)Pr( two working fuses are selected| lot is from B) + P(lot is from A) Pr( two working fuses are selected from lot | lot is from A).

The wording of the problem does not tell us that each bath of fuses from manufacturer A contains exactly 4 defective fuses. So we must treat the probability that a "randomly selected fuse" from manufacturer A is defective as .04. By contrast if we knew that each batch of fuses from A contained exactly 4 defective fuses, we would be faced with a situation of "random sampling without replacement".

The wording of the problem says that a given "lot" of fuses consists of fuses from only one manufacturer, not fuses from both manufacturers. The numbers you used in your solution seem to come from considering fuses taken from both manufacturers during one test for acceptance.

 

[BvU]

So what are P(A) and P(B) ? How do you think this testing takes place ? And how do you explain your 0.99^2 x (0.96^2 + 0.96(0.04) + 0.04^2) + (0.01 (0.99) +0.01^2) x (0.96^2) with this relevant equation ?

 

[Clara Chung]

So what are P(A) and P(B) ? How do you think this testing takes place ? And how do you explain your 0.99^2 x (0.96^2 + 0.96(0.04) + 0.04^2) + (0.01 (0.99) +0.01^2) x (0.96^2) with this relevant equation ?

I interpret the question as taking 2 fuses from each of the box. When the 2 fuses from one box are not defective, the company accept one box. So I thought question a is asking the probability of accepting one box only (not both box accepted nor both not accepted). Now I understand the question is asking the probability of the company accepting one lot when only one box is examined. Thank you so much for both answers!

 

[Ray Vickson]

I interpret the question as taking 2 fuses from each of the box. When the 2 fuses from one box are not defective, the company accept one box. So I thought question a is asking the probability of accepting one box only (not both box accepted nor both not accepted). Now I understand the question is asking the probability of the company accepting one lot when only one box is examined. Thank you so much for both answers!

The wording of (a) asks about a box; that means: (i) we receive a box; (ii) we draw two items from that box; (iii) we accept the box if both tested items are OK. So:
$$P(\text{accept}) = P(A)\, P(\text{accept}|A) + P(B)\, P(\text{accept}|B).$$