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

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Michael F. | Mathematics TutorMathematics Tutor

Marked as Best Answer

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: C_{1} = Box A (call it A), C_{2} = 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

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.

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:

P_{A} = 5/7 [5 even cards in box A ÷ 7 total even cards]

The probability it came for Box B is:

P_{B} = 1 - 5/7 = 2/7 [2 even cards in box A ÷ 7 total even cards]

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