What Is the Probability a Lemon Dreamboat Car Came from Factory A?

[silkdigital]

Question:

Dreamboat cars are produced at three different factories A, B, and C. Factory A produces 20% of the total output of Dreamboats, B 50%, and C 30%. However, 5% of the cars at A are lemons, 2% at B are lemons, 10% at C are lemons. If you buy a Dreamboat and it turns out to be a lemon, what is the probability that it was produced at factory A?

My workings were:

0.05 / (0.05 + 0.02 + 0.1) = 29.41%

I ignored the output percentages because the base condition is that the car is a lemon already. Is my logic on the right track? Not sure if I'm correct!

Another tough question is this:

Suppose that 10 cards, of which 7 are red and 3 are green, are put at random into 10 envelopes, of which seven are red and three are green, so that each envelope contains one card. Determine the probability that exactly k envelopes will contain a card with a matching color.

I've managed to obtain k=4
[7C4 * 3C3] / 10C7

and k=10
[7C7 * 3C0] / 10C7

Not sure what the next steps are to express in terms of any k. Hope that makes sense.
Thanks for any help in advance, having some difficulty!

 

[awkward]

Silkdigital,

For the dreamboats, you need
$$P(\text{produced at A} | \text{is a lemon}) = \frac{P(\text{produced at A and is a lemon)}}{P(\text{is a lemon})}$$
for which I think you need the output percentages.

For the cards, it may be easier to answer a closely related question: How many green cards are in green envelopes? This is exactly like taking a random sample of three cards without replacement and asking how many of them are green.

 

[SW VandeCarr]

Question:

Dreamboat cars are produced at three different factories A, B, and C. Factory A produces 20% of the total output of Dreamboats, B 50%, and C 30%. However, 5% of the cars at A are lemons, 2% at B are lemons, 10% at C are lemons. If you buy a Dreamboat and it turns out to be a lemon, what is the probability that it was produced at factory A?

My workings were:

0.05 / (0.05 + 0.02 + 0.1) = 29.41%

Note, you failed to weight your error percentages by the production percentages. The weights can be assigned in various ways provided the proportions are preserved. So for factory A: (0.2)(0.05)/[(0.2)(0.05)+(0.5)(0.02)+(0.3)(0.10)] = 0.01/(0.01+0.01+0.03)=0.2

So if you have a defect, the probabilities are 0.2 from A, 0.2 from B and 0.6 from C. Note they sum to 1.

 

[silkdigital]

Note, you failed to weight your error percentages by the production percentages. The weights can be assigned in various ways provided the proportions are preserved. So for factory A: (0.2)(0.05)/[(0.2)(0.05)+(0.5)(0.02)+(0.3)(0.10)] = 0.01/(0.01+0.01+0.03)=0.2

So if you have a defect, the probabilities are 0.2 from A, 0.2 from B and 0.6 from C. Note they sum to 1.

Yes I realized it now. I guess I misinterpreted the question. I did the same thing and ended with 0.2, for the envelope question I realized matches only exist for even k when k>=4 (ie 4,6,8,10) and all marginal probabilities add to 1.
Thanks for the help guys!

 

[SW VandeCarr]

Yes I realized it now. I guess I misinterpreted the question. I did the same thing and ended with 0.2, for the envelope question I realized matches only exist for even k when k>=4 (ie 4,6,8,10) and all marginal probabilities add to 1.
Thanks for the help guys!

You're welcome.