Among three indistinguishable boxes one contains two pennies, a second contains one penny and one dime and a third contains two dimes. Suppose we choose a box at random and choose a coin from that box. The coin turns out to be a penny. What is the probability that the other coin in the same box is a penny?

*Please I would like to understand what the formula is and how it works*

2/5 off the too of my head

I need to understand how it's done more than the answer. Any help?

We'll how many pennies do you have total? Subtract 1 because you already chose it.. You have 3 total... You've taken 1 out.. U have 2 left.. So you can chose 2pennies from 3. 2/3 sorry

You should reread the question. It's asking what the probability that the other coin in the same box is a penny?

Now with this in mind, the selected coin was a penny which would leave either the possibility that it's a dime or penny. So isn't it 50 percent?

There are six possible outcomes to selecting a box, a first coin, and then the second coin:

1. Box A, penny 1, penny 2
2. Box A. penny 2, penny 1
3. Box B, penny, dime
4. Box B, dime, penny
5. Box C. dime1, dime 2
6. Box C, dime 2, dime 1

If you pull out a penny first, then you know that the situation 4, 5, and 6 can be ruled out. So given that you have situations 1, 2 and 3 still in play - what's the probability that the second coin is a penny?

You could also approach this using Bayes' theroem: P(A|B) = P(A & B)/P(A), Here we set P(B) = probability of getting a penny on the first draw, P(A) = probability of getting a penny on the second draw, and P(A&B) = probability of gerting a penny on both the first and second draws. So P(A|B) = P(penny on second draw given penny on first draw) = P(penny on both draws)/P(penny on first draw) = (2/6)/(3/6) = 2/3.

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Originally Posted by ebaines

There are six possible outcomes to selecting a box, a first coin, and then the second coin:

1. Box A, penny 1, penny 2
2. Box A. penny 2, penny 1
3. Box B, penny, dime
4. Box B, dime, penny
5. Box C. dime1, dime 2
6. Box C, dime 2, dime 1

If you pull out a penny first, then you know that the situation 4, 5, and 6 can be ruled out. So given that yuo have situations 1, 2 and 3 still in play - what's the probability that teh second coin is a penny?

You could also approach this using Bayes' theroem: P(A|B) = P(A & B)/P(A), Here we set P(B) = probability of getting a penny on the first draw, P(A) = probability of getting a penny on the second draw, and P(A&B) = probability of gerting a penny on both the first and second draws. So P(A|B) = P(penny on second draw given penny on first draw) = P(penny on both draws)/P(penny on first draw) = (2/6)/(3/6) = 2/3.

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Thanks for explaining the bayes' theorem! I've been trying to apply to this but haven't had the proper explanation. Kudos!