# Thread: Selecting balls numbered 1 through 4 from an urn

1. ## Selecting balls numbered 1 through 4 from an urn

An urn contains four balls numbered 1 through 4. The balls are selected one at a time without replacement. A match occurs if ball numbered m is the mth ball selected. Let the event Ai denote a match on the ith draw, i = 1,2,3,4.

a) Show that P(Ai) = 3!/4!

b) Show that P(Ai Intersection Aj) = 2!/4!

c) Show that P(Ai Intersection Aj intersection Ak) = 1!/4!

d) Show that the probability of at least one match is P (A1 U A2 U A3 U A4) = 1 - 1/2! + 1/3! - 1/4!

e) Extend this exercise so that there are n balls in the urn. Show that the probability of at least one match is:

P(A1 U A2 U...U An) = 1 - 1/2! + 1/3! - 1/4! +...+ (-1)n+1 = 1 - (1- 1/1! + 1/2! - 1/3! +...+ (-1)n ).
n! n!

This is exercise 1.4-19 from Probability and Statistical Inference, Hogg and Tanis 8th edition. If anyone could help me with this problem I would greatly appreciate it.

2. ## Re: Selecting balls numbered 1 through 4 from an urn

You are supposed to show some work.

Originally Posted by polcareb192
a) Show that P(Ai) = 3!/4!
Of all permutations of 4 elements, you select those that leave i in place. The other 3 elements can come in an arbitrary order.

Originally Posted by polcareb192
b) Show that P(Ai Intersection Aj) = 2!/4!
Provided that i ≠ j. This is similar to a).

Originally Posted by polcareb192
c) Show that P(Ai Intersection Aj intersection Ak) = 1!/4!
Provided i, j and k are pairwise distinct.

Originally Posted by polcareb192
d) Show that the probability of at least one match is P (A1 U A2 U A3 U A4) = 1 - 1/2! + 1/3! - 1/4!
Use the inclusion-exclusion principle.

3. ## Re: Selecting balls numbered 1 through 4 from an urn

***The underlined terms should be over n!***

4. ## Re: Selecting balls numbered 1 through 4 from an urn

Originally Posted by polcareb192
An urn contains four balls numbered 1 through 4. The balls are selected one at a time without replacement. A match occurs if ball numbered m is the mth ball selected. Let the event Ai denote a match on the ith draw, i = 1,2,3,4.

a) Show that P(Ai) = 3!/4!
Why in the world write it in such a strange way? 3!/4!= 1/4. And that's easy to show.

A1 is the event of getting ball number 1 on the first draw: there are 4 balls in the urn so the probability of drawing any specific ball on the first draw is 1/4.

A2 is the event of getting ball number :2" on the second draw. Since this is "sampling without replacement" in order to get ball "2" on the second draw we must NOT have got it on the first draw. There were 4 balls, 3 of which were not labled "2" so the probability of getting anything except "2" on the first draw is 3/4. Once we have done that, there are 3 balls left, one of which is labeled "2". The probability of getting that ball on the second draw is 1/3 so the probability of getting "ball number "2" on the second draw" is (3/4)(1/3)= 1/4.

Similarly, to get ball "3" on the third draw we must NOT get it on the first draw, which has probability 3/4, then not get it on the second draw, which has probability 2/3, then get it on the third draw, which has probability 1/2. The probability of getting "ball "3" on the third draw" is (3/4)(2/3)(1/2)= 1/4.

Of course, the probability of getting the ball marked "4" on the fourth draw is (3/4)(2/3)(1/2)(1/1)= 1/4.

b) Show that P(Ai Intersection Aj) = 2!/4!
2!/4!= 1/12 and can be argued in a similar manner to (a).

c) Show that P(Ai Intersection Aj intersection Ak) = 1!/4!

d) Show that the probability of at least one match is P (A1 U A2 U A3 U A4) = 1 - 1/2! + 1/3! - 1/4!

e) Extend this exercise so that there are n balls in the urn. Show that the probability of at least one match is:

P(A1 U A2 U...U An) = 1 - 1/2! + 1/3! - 1/4! +...+ (-1)n+1 = 1 - (1- 1/1! + 1/2! - 1/3! +...+ (-1)n ).
n! n!

This is exercise 1.4-19 from Probability and Statistical Inference, Hogg and Tanis 8th edition. If anyone could help me with this problem I would greatly appreciate it.

5. ## Re: Selecting balls numbered 1 through 4 from an urn

Originally Posted by HallsofIvy
Why in the world write it in such a strange way? 3!/4!= 1/4. And that's easy to show.
I think that the problem author computes the probabilities of events by definition: the cardinality of an event divided by the cardinality of the sample space. (It is assumed that all singletons in the sample space have equal probability.) The sample space here consists of permutations of (1, 2, 3, 4). The event Ai consists of all permutations where the ith element is i, and other elements are arbitrary. Thus, |Ai| = 3!. My guess is that 3! / 4! gives a hint, which is described above, how to compute the answer 1 / 4. This method is arguable easier: it finds P(Ai) for all i at once and it does not use conditional probability.

6. ## Re: Selecting balls numbered 1 through 4 from an urn

Originally Posted by polcareb192
An urn contains four balls numbered 1 through 4. The balls are selected one at a time without replacement. A match occurs if ball numbered m is the mth ball selected. Let the event Ai denote a match on the ith draw, i = 1,2,3,4.

a) Show that P(Ai) = 3!/4!

b) Show that P(Ai Intersection Aj) = 2!/4!

c) Show that P(Ai Intersection Aj intersection Ak) = 1!/4!

d) Show that the probability of at least one match is P (A1 U A2 U A3 U A4) = 1 - 1/2! + 1/3! - 1/4!

e) Extend this exercise so that there are n balls in the urn. Show that the probability of at least one match is:

P(A1 U A2 U...U An) = 1 - 1/2! + 1/3! - 1/4! +...+ (-1)n+1 = 1 - (1- 1/1! + 1/2! - 1/3! +...+ (-1)n ).
n! n!

This is exercise 1.4-19 from Probability and Statistical Inference, Hogg and Tanis 8th edition. If anyone could help me with this problem I would greatly appreciate it.

ok i was working on this problem and this form wasn't that helpfull so once i figured it out i decided to address the problem on here

a) for the right ball to the right slot is a 1/4 chance the people above crunched the numbers but it is kinda easy to tell if you think of this like the urn problems... you have 1 Ai and 3 non-Ais
but we beed to permute them for all possible outcomes... sooooo 1/4 X 4!/(1!1!1!1!) gets you how many posible there are (namly 3!)
take that and divide by the total number of possibilities (namly 4!)

b) just like a we have 1/4 for the first match, but for the second it is 1/3 because there is only 3 total balls left.... then we have to shuffle them all so we get (1/4)(1/3)(4!/1!1!1!1!)

c) simularly is (1/4)(1/3)(1/2)(4!/1!1!1!1!)

d) the inclustion exclusion formula works out for this... but what is interesting to note is that you have [ ∑(from x+1 to 4) (4 choose x)(p(Ai U......U Ax) , {where 0 i ≤ x} ] [So x is the number of matches]

e) this is where that explanation in d) matters... the formula turns into [∑(from x+1 to 4) (n choose x)(p(Ai U......U Ax) , {where 0 i ≤ x} ]

note: (n choose x)(p(Ai U......U Ax) , {where 0 i ≤ x} is equal to 1/x!

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# an urn contains 4 balls numbered 1 through 4

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