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Probability question

Hi, I have a question on probability that I don't know how to approach:
 
Box A contains 10 cards numbered 1 through 10, and box B contains 5 cards numbered 1 through 5. A box is selected at random and a card is randomly drawn from the box. If the number is even, find the probability the card came from box A.
 
Thank you

Comments

In my answer, change the notation "EBD" to "ECD"  (cards, not balls!).  Sorry for the slip.
 
Dattaprabhakar (Dr. G.)
Ann:
 
To really know whether or not you fully understand how the problem is solved, I would like you to do two more exercises.  They should not take more tham 15 minutes of your time.
 
Exercise 1: Suppose that a box (either A or B) is selected with a probability proportional to card it contains.  Everything else in the problem remains the same.  A card is randomly drawn from the box. If the number drawn is even, find the probability that the card came from box A.
 
Exercise 2.  Go back to the problem you asked.  Now add another box, say Box C, with 4 cards numbered 1 to 4.  A card is randomly drawn from the box.If the number drawn is even, find the probability that the card came from box A.
 
I will be happy if you post a comment with your answers and doubly happy if they are correct!  (I would have given you a small reward, if that was possible.) Of course, I will post the correct answers if you want to check.
 
Dattaprabhakar ("Dr. .")
Suppose that a box (either A or B) is selected with a probability proportional to card it contains. Is this really ???
Suppose that a box (either A or B) is selected with a probability proportional to the number of cards it contains.
If not, please clarify.
Hi Michael F:
This exercise changes the probabilities associated with A and B.  Selecting the two boxes at random implies P(A) = P(B) = 1/2.  On the other hand, if they are selected with probability proportional to the cards they contain, P(A) is proportional to 10 and P(B) is proportional to 5. Since there are only two boxes P(A) + P(B) = 1. Hence P(A) = 10/15 = 2/3 amd P(B) = 1/3.   These values can now be used to get P(ECD|A).  (see my solution. You will get a different answer.
 
Please post a comment if it is clear now.  Thanks.  I appreciate your interest.
 
Dattaprabhakar ("Dr. G.")
 
Hi Dattaprabhakar G. 
I thought that "proportional to the number of cards it contains" is a clearer statement of your intent than is "proportional to the cards it contains".  We are in agreement.
Hi Michael F.:
 
Thanks.  I also would like to request you read another probability problem ("One billion songs..."),  go over my solution and comment.  Thanks in advance for taking the time.
 
Dattaprabhakar ("Dr. G,")
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4 Answers

Let E=set of even numbered cards
E=(E∩A)∪(E∩B) a disjoint union
P(E)=P(E∩A)+P(E∩B)=5/10+2/5=5/10+4/10=9/10!!!
P(A|E)=P(A∩E)/P(E)=(5/10)/(9/10)=5/9
Sorry for the arithmetic error. Same theorem, different statement.
Application of Bayes Theorem:
 
Conditions: C1 = Box A (call it A),  C2 = Box B (call it B)
 
There are 5 even cards in A out of 10  and 2 even cards in B out of 5.
 
Outcome: Even Card Drawn  (call the outcome "EBD")
 
Want P(A | EBD)
 
By Bayes Theorem,
 
P(A | EBD) = P( EBD| A) P(A) / {P( EBD | A) P(A) + P(EBD|B) P(B)}
 
                 = (5/10)(1/2) / {(5/10)(1/2) + (2/5) (1/2)}
                   = 5 / 9
 

Comments

To all interested:
 
Bayes Theorem updates current knowledge of a system in the light of an observed outcome. Suppose that there are some conditions in the system that currently are known to occur with known probabilities. Each condition gives rise to many outcomes, also with known probabilities. In the light of the fact a certain outcome has occurred, Bayes Theorem allows us to update the probability of a given condtion that would give rise to that particular outcome.

In our illustrative problem, the system is drawing randomly a numbered card from boxes and noting whether it is even or odd. The three “conditions” are, let us say, three boxes, Box A with 10 cards numbered 1 to 10, Box B with 5 cards numbered 1 to 5 and Box C with 4 cards numbered 1 to 4. Let us begin by assuming that the conditions occur at random, so that probabilities associated with them are
P(A) = P(B) = P(C) = 1/3.
Under each condition there are two outcomes, “even” (“E”) or “odd” (“O”). The known probabilities of their occurrence under each condition are
P(E|A) = 5/10, P(O|A) = 5/10, P(E|B) = 2/5, P(O|B) = 3/5, P(E|C) = 2/4, P(O|C) = 2/ 4.

Now suppose that we observe the outcome (only) that an even ball is drawn (E has occurred). How do you use this information to update that Box A, or Box B, or Box C is used (is the prevalent condition)? So we now want P(A|E), P(B|E) and P(C|E) (the updated probabilities of the conditions, in the light of the occurrence of outcome E). Let us take P(A|E).

By Bayes Theorem,

P(A|E) = [P(E|A)P(A)] / {[P(E|A)P(A)] + [P(E|B)P(B)] + [P(E|C)P(A|C)]}

All the quantities on the right are known (see above) so P(A|E) can be computed.

P(A|E) = (5/10)(1/3) / {(5/10)(1/3) + (2/5)(1/3) + (2/4)(1/3)}

Remaining arithmetic is left to the interested reader.

Same goes for P(B|E) and P(C|E). Just replace A in the expression above throughout by B,  or by C.
 
This powerful technique of Bayes Theorem can be analogously used in many important real life situations.
 
Dattaprabhakar ("Dr. G.")
There are 5 even cards in Box A (2, 4, 6, 8, 10) and 2 even Cards in Box B (2, 4), for a total of 7 even cards.  The probability of the even card coming from Box A is:
 
PA = 5/7                    [5 even cards in box A ÷ 7 total even cards]
 
The probability it came for Box B is:
 
PB = 1 - 5/7 = 2/7     [2 even cards in box A ÷ 7 total even cards]
     I'm feeling that the answer is 1/4 because you are doing 2 activities at random, picking a box and picking an even card. so the probability that you pick box A is 1/2 and the probability you pick an even from box A is 1/2. Multiply those together and you get 1/4 .

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