# Confusion regarding a probability problem

A certain auditorium has 30 rows of seats. Row 1 has 11 seats, while Row 2 has 12 seats, Row 3 has 13 seats, and so on to the back of the auditorium where Row 30 has 40 seats. A door prize is to be given away by randomly selecting a row (with equal probability of selecting any of the 30 rows) and then randomly selecting a seat within that row (with each seat in the row equally likely to be selected). Now, 1) Find the probability that Seat 15 was selected given that row 20 was selected ? 2) Find the probability that Row 20 was selected given that Seat 15 was selected ?

To answer the first, given that Row 20 was selected, there are 30 possible seats in Row 20 that are equally likely to be selected. Hence Pr(Seat 15 | Row 20) = 1/30. The same kind of argument can be given to answer the second : given that Seat 15 was selected, there are 30 possible rows that are equally likely to be selected. Hence Pr(Row 20 | Seat 15) = 1/30.

Now, it turns out that the first answer is correct whereas the second answer is incorrect. My question is where am I making mistakes in computing the second answer ?

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First comment: Some of the rows have fewer than 15 seats, so if seat 15 was selected then there aren't 30 possible rows. Second comment: Even if you use the correct number of rows, I don't think your approach will give the correct answer. Try using the definition of conditional probability. –  mark999 Oct 7 '12 at 9:07
To add to mark999's comment, you need to use Bayes' formula: $$P(\text{row}~20\mid\text{seat}~15)=\frac{P(\text{seat}~15\mid\text{seat}~20)P(‌​\text{seat}~20)}{P(\text{seat}~15)}$$ where the denominator on the right is computed via the law of total probability: $$P(\text{seat}~15)=\sum_{i=11}^{40}P(\text{seat}~15\mid\text{row}~i)P(\text{row‌​}~i).$$ Note that some of the terms on the right are $0$, $P(\text{row}~i)$ is the same for all $i$, and the numerator in Bayes' formula is one of the terms in the sum used in computing the denominator. –  Dilip Sarwate Oct 7 '12 at 12:52
@DilipSarwate: Typo: In numerator, "seat 20" should read "row 20". –  Arthur Small Dec 7 '12 at 1:59
@ArthurSmall You are correct. Thanks for pointing it out. Unfortunately, it is too late to correct the comment. –  Dilip Sarwate Dec 7 '12 at 2:59

## migrated from math.stackexchange.comOct 7 '12 at 23:13

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"given that Seat 15 was selected, there are 30 possible rows that are equally likely to be selected". This is incorrect; the rows are not equally likely to be selected. Conditioned on a particular seat number having been selected, it's more likely to have come from a smaller row with less alternatives.

For example, suppose there were only two rows, the first with just a Seat #1, the second with a Seat #1 and a Seat #2. 50% of the time, you pick Row #1 and 50% of the time, you pick Row #2. And within Row #2, you pick Seat #1 and Seat #2 equally as often. Then Row 1 Seat 1 gets chosen 50% of the time overall, Row 2 Seat 1 gets chosen 25% of the time overall, and Row 2 Seat 2 gets chosen 25% of the time overall. Which means, out of all the times Seat #1 gets chosen, it is in Row 1 twice as often as it is in Row 2. Pr(Row 1 | Seat 1) = 50%/(50% + 25%) = 2/3, rather than 1/2.

That same phenomenon is occurring here.

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Thanks for your answer. Can we make any general statement from these using "equally probable": when and when not to use Bayes's theorem to calculate conditional probability ? –  Prasenjit Oct 7 '12 at 6:59
@Prasenjit: Always. It's just that sometimes we are made to take shortcuts. :) By the way you should at least have noticed that row 1 cannot possibly have been selected if seat 15 is. –  Hagen von Eitzen Oct 7 '12 at 9:34

Let me put above answer in more simple way: (2nd case)

Standard formula for conditional probability is:
P(A|B) = P(A∩B)/P(B)

Let A is event of Row#20 selected
and B is event of Seat#15 selected

So, P(A∩B) = 1/30 * 1/30 (As 30 rows, and 30 seats in row#20)

and P(B) = SUM {for_each_row}: {prob_of_selecting_row}*{prob_of_B_in_that_row}.

As prob. of selecting any row is same for each row,
P(B) = 1/30 * (1/15 + 1/16 + ... + 1/40)

So finally, P(A|B) = (1/30) / (1/15 + 1/16 +... + 1/40)
Done!

• What was the mistake?

Selection of seat# is dependent on row selected, so selection of row# is not common for all row. You made mistake by taking P(Row#15)=1/30 for all possible row.

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