# Probability Problem

• Apr 21st 2008, 04:20 PM
desemejante
Probability Problem
I'm actually completely lost on this one...just cannot get my head around the probability thing:

An examiner gives different marks to each of three scripts that she
corrects. The names of the three candidates are not on the scripts; and she
has lost the list associating their names with the numbers on the scripts;
hence she assigns the three marks randomly to the three candidates. What
is the probability that:
(i) all three candidates get the correct mark?
(ii) none do?
(iii) at least two do?
(iv) exactly two do?
(v) only one does?
(vi) at least one does?

Thanks so much!
• Apr 21st 2008, 04:54 PM
mr fantastic
Quote:

Originally Posted by desemejante
I'm actually completely lost on this one...just cannot get my head around the probability thing:

An examiner gives different marks to each of three scripts that she
corrects. The names of the three candidates are not on the scripts; and she
has lost the list associating their names with the numbers on the scripts;
hence she assigns the three marks randomly to the three candidates. What
is the probability that:
(i) all three candidates get the correct mark?
(ii) none do?
(iii) at least two do?
(iv) exactly two do?
(v) only one does?
(vi) at least one does?

Thanks so much!

Let X be the random variable number of scripts that get the correct mark.

X ~ Binomial(n = 3, p = 1/3).

If you're not familiar with the binomial distribution, please say so and I'll show how to get these from first principles using counting (combinatorial)techniques.

(i) Pr(X = 3) = ....
(ii) Pr(X = 0) = ....
(iv) Pr(X = 2) = ....
(iii) $\Pr(X \geq 2) = \Pr(X = 2) + \Pr(X = 3) = ....$ (Be smart here - use the answers from i and iv)
(v) Pr(X = 1) = .....
(vi) 1 - Pr(X = 0) = .....
• Apr 21st 2008, 05:21 PM
desemejante
Thanks so much! I know the binomial distribution, should have thought of this...
• Apr 21st 2008, 07:02 PM
awkward
Quote:

Originally Posted by desemejante
I'm actually completely lost on this one...just cannot get my head around the probability thing:

An examiner gives different marks to each of three scripts that she
corrects. The names of the three candidates are not on the scripts; and she
has lost the list associating their names with the numbers on the scripts;
hence she assigns the three marks randomly to the three candidates. What
is the probability that:
(i) all three candidates get the correct mark?
(ii) none do?
(iii) at least two do?
(iv) exactly two do?
(v) only one does?
(vi) at least one does?

Thanks so much!

desemenjante,

It all hinges on your interpretation of the phrase "she assigns the three marks randomly to the three candidates". Let's say the marks are 70, 80, and 90. If you think she simply chooses a mark at random from the list for each candidate, (then for example, all three candidates might be assigned a mark of 70), you will end up with a binomial distribution for the number of correct matchings, as suggested by Mr. Fantastic.

However, an alternative interpretation that seems more likely to me is that she randomly selects one of the 3! = 6 permutations of the marks. So, for example, the three candidates might be assigned marks of 80, 70, and 90, but it is not possible that all three candidates might be assigned the same mark. With this interpretation you will not get a binomial distribution for the number of correct matchings, because the matchings are not independent. The answer to (i) is then 1/6, because only one of the 6 permutations leads to a correct match for all three candidates. (I'm assuming here that the marks are distinct, otherwise you will have a different problem.) To answer the other questions, the simplest strategy is to list all 6 permutations and count the number of correct matches in each case.
• Apr 21st 2008, 10:31 PM
mr fantastic
Quote:

Originally Posted by awkward
desemenjante,

It all hinges on your interpretation of the phrase "she assigns the three marks randomly to the three candidates". Let's say the marks are 70, 80, and 90. If you think she simply chooses a mark at random from the list for each candidate, (then for example, all three candidates might be assigned a mark of 70), you will end up with a binomial distribution for the number of correct matchings, as suggested by Mr. Fantastic.

However, an alternative interpretation that seems more likely to me is that she randomly selects one of the 3! = 6 permutations of the marks. So, for example, the three candidates might be assigned marks of 80, 70, and 90, but it is not possible that all three candidates might be assigned the same mark. With this interpretation you will not get a binomial distribution for the number of correct matchings, because the matchings are not independent. The answer to (i) is then 1/6, because only one of the 6 permutations leads to a correct match for all three candidates. (I'm assuming here that the marks are distinct, otherwise you will have a different problem.) To answer the other questions, the simplest strategy is to list all 6 permutations and count the number of correct matches in each case.

You're right. Good save.
• Apr 22nd 2008, 01:23 AM
desemejante

The last one is te one I had in mind...but "counting" combinations seemed so "unmathematical" and completely unlike my tutor ;)

He made us do another great one:

In a certain country, all married couples desire 2 male children. Up to a maximum of
n = 4 children, they continue to procreate until they have two boys. Male and female children are equally probable and successive sexes
are independent. What is (i) the average family size and (ii) the sex ratio,
in this country? How would these numbers change if
n could be infinitely
large?

I'm confused about whether they'd try to have a fourth child if they already have three girls (considering they cannot make their "2 male children" goal then anyway.

• Apr 22nd 2008, 03:13 AM
CaptainBlack
Quote:

Originally Posted by desemejante
I'm actually completely lost on this one...just cannot get my head around the probability thing:

An examiner gives different marks to each of three scripts that she
corrects. The names of the three candidates are not on the scripts; and she
has lost the list associating their names with the numbers on the scripts;
hence she assigns the three marks randomly to the three candidates. What
is the probability that:
(i) all three candidates get the correct mark?
(ii) none do?
(iii) at least two do?
(iv) exactly two do?
(v) only one does?
(vi) at least one does?

Thanks so much!

(iv) is impossible so has probability 0.

As (iv) is impossible (iii) is the same as (i)

RonL
• Apr 22nd 2008, 03:52 AM
mr fantastic
Quote:

Originally Posted by desemejante

The last one is te one I had in mind...but "counting" combinations seemed so "unmathematical" and completely unlike my tutor ;)

He made us do another great one:

In a certain country, all married couples desire 2 male children. Up to a maximum of
n = 4 children, they continue to procreate until they have two boys. Male and female children are equally probable and successive sexes

are independent. What is (i) the average family size and (ii) the sex ratio,

in this country? How would these numbers change if n could be infinitely

large?

I'm confused about whether they'd try to have a fourth child if they already have three girls (considering they cannot make their "2 male children" goal then anyway.

Here's the way I see it:

(i) Let X be the random variable number of children in family.

X = 2: BB. Pr(X = 2) = 1/4.

X = 3: BGB or GBB. Pr(X = 3) = 1/8 + 1/8 = 1/4.

X = 4: GBGA, GGBA or GGGA where A is either B or G. Pr(X = 4) = 1 - 1/4 - 1/4 = 1/2.
(I'm assuming that families have the maximum possible number of children, regardless of whether they get two boys or not).

E(X) = (2)(1/4) + 3(1/4) + 4(1/2) = 13/4.

(ii) Let Y be the random variable number of boys in a family.

Y = 2:
BB. Pr(Y = 2 | X = 2) = 1/4.
BGB or GBB. Pr(Y = 2 | X = 3) = 1/8 + 1/8 = 1/4.
GBGB, GGBB or BGGB. Pr(Y = 2 | X = 4) = 3/16.

Therefore Pr(Y = 2) = 11/16.

Y = 1:
BGGG, GBGG, GGBG or GGGB. Pr(Y = 1 | X = 4) = 1/4.

Therefore Pr(Y = 1) = 1/4.

Y = 0:
GGGG. Pr(Y = 0 | X = 4) = 1/16.

Therefore Pr(Y = 0) = 1/16.

E(Y) = (0)(1/16) + (1)(1/4) + (2)(11/16) = 26/16 = 13/8.

Since E(X) = 13/4, the expected number of girls is 13/4 - 13/8 = 13/8.

This makes sense since, if you let Z be the random variable number of girls in a family:

Pr(Z = 0) = Pr(X = 2) = 1/4.
Pr(Z = 1) = Pr(X = 3) = 1/4.
Pr(Z = 2) = 3/16. (GBGB, GGBB or BGGB).
Pr(Z = 3) = Pr(Y = 1) = 1/4.
Pr(Z = 4) = 1/16. (GGGG)

E(Z) = (0)(1/4) + (1)(1/4) + (2)(3/16) + (3)(1/4) + (4)(1/16) = 13/8.

So I get the ratio to be 13/8 : 13/8 <=> 1:1.

(iii) How would these numbers change if n could be infinitely large?

Let X be the random variable number of children until the first boy.
Let Y be the random variable number of children after first boy until the second boy.

The X and Y both follow a geometric distribution with parameter p = 1/2. Note that E(X) = E(Y) = 1/p = 2.

Let Z = X + Y.

E(Z) = E(X) + E(Y) = 2 + 2 = 4.

I'll leave the ratio question for you to consider.
• Apr 23rd 2008, 11:56 AM
desemejante
Quote:

Originally Posted by mr fantastic
(ii) Let Y be the random variable number of boys in a family.

Y = 2:
BB. Pr(Y = 2 | X = 2) = 1/4.
BGB or GBB. Pr(Y = 2 | X = 3) = 1/8 + 1/8 = 1/4.
GBGB, GGBB or BGGB. Pr(Y = 2 | X = 4) = 3/16.

Therefore Pr(Y = 2) = 11/16.

Y = 1:
BGGG, GBGG, GGBG or GGGB. Pr(Y = 1 | X = 4) = 1/4.

Therefore Pr(Y = 1) = 1/4.

Y = 0:
GGGG. Pr(Y = 0 | X = 4) = 1/16.

Therefore Pr(Y = 0) = 1/16.

E(Y) = (0)(1/16) + (1)(1/4) + (2)(11/16) = 26/16 = 13/8.

Since E(X) = 13/4, the expected number of girls is 13/4 - 13/8 = 13/8.

This makes sense since, if you let Z be the random variable number of girls in a family:

Pr(Z = 0) = Pr(X = 2) = 1/4.
Pr(Z = 1) = Pr(X = 3) = 1/4.
Pr(Z = 2) = 3/16. (GBGB, GGBB or BGGB).
Pr(Z = 3) = Pr(Y = 1) = 1/4.
Pr(Z = 4) = 1/16. (GGGG)

E(Z) = (0)(1/4) + (1)(1/4) + (2)(3/16) + (3)(1/4) + (4)(1/16) = 13/8.

So I get the ratio to be 13/8 : 13/8 <=> 1:1.