Is this probability right?

• Mar 8th 2009, 05:30 AM
wallace
Is this probability right?
Ok
Question 1
---------------
A factory has two production lines for microwave ovens, A and B. The probability that a microwave oven from line A will prove to be faulty when checked is 0.008. The probability that a microwave oven from line B will prove to be faulty is 0.012. Line A produces three times as many microwave ovens as line B.
(i)
Find the probability that a randomly selected microwave oven will prove to be faulty.

Probability A or B to be faulty = 0.008 x 0.012
Probability A or B to be faulty = 0.0096

My doubt!
Do I need to consider that Line A produces 3 times more than B??? Even knowing that the probability is 0.008 for the total production??

(ii)
If a microwave oven is found to be faulty, find the probability that it was produced on line A.

Here i did -

P(not B) = P(A) or P(B) / P(B) = 0.00096 x 0.012 = 0.08
• Mar 8th 2009, 05:56 AM
HallsofIvy
Quote:

Originally Posted by wallace
Ok
Question 1
---------------
A factory has two production lines for microwave ovens, A and B. The probability that a microwave oven from line A will prove to be faulty when checked is 0.008. The probability that a microwave oven from line B will prove to be faulty is 0.012. Line A produces three times as many microwave ovens as line B.
(i)
Find the probability that a randomly selected microwave oven will prove to be faulty.

Probability A or B to be faulty = 0.008 x 0.012
Probability A or B to be faulty = 0.0096

My doubt!
Do I need to consider that Line A produces 3 times more than B??? Even knowing that the probability is 0.008 for the total production??

Yes, you do. Imagine that there were 4000 microwave ovens produced. Then, since A produces 3 times as many (NOT "three times more"- that would be 4 times as many) as B, 3000 of those ovens are from A and 1000 from B. Since the probability that that ovens from A are defective is .008, there would be .008(3000)= 24 defective ovens from A. Since the probability that ovens from B are defective is 0.012, there will be 0.012(1000)= 12 defective ovens from B. That is, a total of 12+24= 36 defective ovens out of 4000: the overall probability that an oven will be defective is 36/4000= 9/1000= 0.009.

By the way, your basic formula is wrong. P(A)*P(B) is "probability of A and B" if A and B are independent events, not "probability of A or B". That is P(A)+ P(B)- P(A and B).

Quote:

(ii)
If a microwave oven is found to be faulty, find the probability that it was produced on line A.

Here i did -

P(not B) = P(A) or P(B) / P(B) = 0.00096 x 0.012 = 0.08
I have no idea what you are doing there. "P(not B)" is NOT P(A) or P(B)/P(B), it is 1- P(B) but, in any case, that has nothing to do with the conditional probability asked for here.

Look at my example above. There were a total of 36 defective ovens, 24 of them from A. If an oven is defective, the probability it was from A is 24/36= 2/3.
• Mar 8th 2009, 06:28 AM
wallace
Thanks so much!

So Would it be like this??

i) (P A and B) = 0.0096

Probability of a microwave to be faulty is = p(a) + p(b) - (p and b)

P(faulty) = 0.008 + 0.012 - 0.0096

P(faulty) = 0.0104

Correct??

ii) I need to found the probabilty that it was produced in line A.
I think I need to use baye's formula that would be

p(a|b) = P(B|A)P(A) / P(B)

• Mar 8th 2009, 02:45 PM
mr fantastic
Quote:

Originally Posted by wallace
First question I think I knew how to do. please confirm if its right!

Question 1
---------------
A factory has two production lines for microwave ovens, A and B. The probability that a microwave oven from line A will prove to be faulty when checked is 0.008. The probability that a microwave oven from line B will prove to be faulty is 0.012. Line A produces three times as many microwave ovens as line B.
(i)
Find the probability that a randomly selected microwave oven will prove to be faulty.

Probability A or B to be faulty = 0.008 x 0.012
Probability A or B to be faulty = 0.0096

My doubt!
Do I need to consider that Line A produces 3 times more than B??? Even knowing that the probability is 0.008 for the total production??

[snip]

Of course. Pr(choosing A) = 3/4 and Pr(choosing B) = 1/4. Now try drawing a tree diagram to answer the questions.

(Would you hesitate to consider this if you were told that line A produces 1,000,000 times more than line B ....?)
• Mar 8th 2009, 03:18 PM
mr fantastic
Quote:

Originally Posted by wallace
Ok
Question 1
---------------
A factory has two production lines for microwave ovens, A and B. The probability that a microwave oven from line A will prove to be faulty when checked is 0.008. The probability that a microwave oven from line B will prove to be faulty is 0.012. Line A produces three times as many microwave ovens as line B.
(i)
Find the probability that a randomly selected microwave oven will prove to be faulty.

Probability A or B to be faulty = 0.008 x 0.012
Probability A or B to be faulty = 0.0096

My doubt!
Do I need to consider that Line A produces 3 times more than B??? Even knowing that the probability is 0.008 for the total production??

(ii)
If a microwave oven is found to be faulty, find the probability that it was produced on line A.

Here i did -

P(not B) = P(A) or P(B) / P(B) = 0.00096 x 0.012 = 0.08