Probability theory

[Bubi]

I need help with this question.
Based on the info in this table, if we randomly select one book, what is the probability of the book being Hardback AND from shelf 2?
Now I know the answer is obvious looking at the table, P(H and S2)=8/71=0.11 but what I find confusing is using the formula for probability of A and B which is:
P(A and B)= P(A) * P(B) . Now when I try to use this formula to solve my question, I get a different result, why is that?
This is what I get : P(Hardback AND shelf2)= P(Hardback) * P(shelf2)= 0.62 * 0.28 = 0.17.

	Hardback	Paperback
shelf 1	11	8
shelf 2	8	12
shelf 3	16	4
shelf 4	9	3

 

[HallsofIvy]

P(A and B)= P(A)*P(B) is true only if A and B are independent events. Here they are not because the total number of hard back and paper back books on shelf 2 are not the same.

You need P(A and B)= P(A)P(B|A) or P(A and B)= P(B)P(A|B).
(P(A|B) is the probability that A is true given that B is true and P(B|A) is the probability that B is true given that A is true.)

Here if a book is hard back, then the probability that it is on shelf 2 is 8/44 and the probability that a book is hardback is 44/71 so the probability a book is hardback and on shelf 2 is (8/44)(44/71)= 8/71.

Or we can say that if a book is on shelf 2 then the probability it is hardback is 8/20 and the probability a book is on shelf 2 is 20/71 so the probability a book is hardback and on shelf 2 is (8/20)(20/71)= 8/71 yet again.

 

[pka]

I need help with this question.
Based on the info in this table, if we randomly select one book, what is the probability of the book being Hardback AND from shelf 2?
Now I know the answer is obvious looking at the table, P(H and S2)=8/71=0.11 but what I find confusing is using the formula for probability of A and B which is:
P(A and B)= P(A) * P(B) . Now when I try to use this formula to solve my question, I get a different result, why is that?
This is what I get : P(Hardback AND shelf2)= P(Hardback) * P(shelf2)= 0.62 * 0.28 = 0.17.

\(\begin{array}{*{20}{c}} {shelf}&{hardback}&{paperback}&{total} \\
\hline 1&{11}&8&{19} \\ 2&{\color{blue}8}&{12}&{20} \\ 3&{16}&4&{20} \\ 4&9&3&{12} \\
\hline {}&{44}&{27}&{\color{blue}71} \end{array}\)
Look at the two blue entries: there are eight hardbacks on shelf 2 out of a total of 71.
This is basically a counting question.

 

[Dr.Peterson]

I need help with this question.
Based on the info in this table, if we randomly select one book, what is the probability of the book being Hardback AND from shelf 2?
Now I know the answer is obvious looking at the table, P(H and S2)=8/71=0.11 but what I find confusing is using the formula for probability of A and B which is:
P(A and B)= P(A) * P(B) . Now when I try to use this formula to solve my question, I get a different result, why is that?
This is what I get : P(Hardback AND shelf2)= P(Hardback) * P(shelf2)= 0.62 * 0.28 = 0.17.

	Hardback	Paperback
shelf 1	11	8
shelf 2	8	12
shelf 3	16	4
shelf 4	9	3

As has been said, the formula you state doesn't apply; the formula that does apply is considerably more complicated, requiring you to do all the counting and a little more. That's why pka has told you to do it just the way you already did!

Lesson: No one in their right mind would bother to use a formula for this problem. Formulas are not always better than just applying a definition and common sense.

 

[HallsofIvy]

As has been said, the formula you state doesn't apply; the formula that does apply is considerably more complicated, requiring you to do all the counting and a little more. That's why pka has told you to do it just the way you already did!

Lesson: No one in their right mind would bother to use a formula for this problem. Formulas are not always better than just applying a definition and common sense.

You expect people to think?!