# Two people draw 3 balls each out of a bag of 13 balls. Various conditions apply. What are the outcomes?

A bag contains 13 balls. Person $A$ chooses randomly 3 balls out of the bag, afterwards Person $B$ does the same.

• Let $E_A$ be the event where $A$ chooses 3 balls of the same colour.

• Let $E_B$ be the event where $B$ chooses 3 balls of the same colour.

• Let $F_A$ be the event where $A$ chooses exactly 2 balls of the same colour.

• Let $F_B$ be the event where $B$ chooses exactly 2 balls of the same colour.

• Let $G$ be the event where $A$ and $B$ have exactly the same selection.

Consider:

a) the balls have different colours;

b) 5 balls are blue, 5 balls are red and 3 balls are green

How many different outcome are possible if:

i) $A$ places the balls back before $B$ chooses.

ii) $A$ does not replace the balls.

1. $E_A$
2. $E_AF_B$
3. $(E_A \bigcup E_B)^c(F_A \bigcup F_B)^c$
4. $F_AF_BG$
5. $F_A \bigcup G^c$
6. $E_A E_B G$

# My problem:

I have an old memo but I don't agree with the solutions.

Their solution for:

• (a.i.5)

$| F_A \bigcup G^c | = {13 \choose 3}{13 \choose 3} = 81796$

My problem: If $G^c$ means $A$ and $B$ have different selections then surely we count ${13 \choose 3}$ too much as somewhere $B$ will have the same ordering as $A$. As such we have to subtract this amount from the above solution they got?

• (b.i.1)

$| E_A | = [{5 \choose 3} +{5 \choose 3} + {3 \choose 3}]{13 \choose 3}$

My problem: I don't get the multiplication with ${13 \choose 3}$

• b.i.2. and b.ii.2. their solution respectively

$| E_A F_B | = [ {5 \choose 3} + {5 \choose 3 } +{3 \choose 3}][{5 \choose 2}{8 \choose 1}+{5 \choose 2}{8 \choose 1} + {3 \choose 2}{10 \choose 1}]$

$| E_AF_B | = {5 \choose 3}[ {2 \choose 2} {8 \choose 1 } + {5 \choose 2} {5 \choose 1} + {3 \choose 2} {7 \choose 1}] + {5 \choose 3}[ {5 \choose 2} {5 \choose 1} + {2 \choose 2} {8 \choose 1 } + {3 \choose 2} {7 \choose 1}] + {3 \choose 3}[ {5 \choose 2} {5 \choose 1} + {5 \choose 2}{5 \choose 1} ]$

My problem: Does $E_A F_B$ not form disjointed sets and therefore $| E_A F_B |$ = 0 for both the above cases?

## ++++++++++ ADDITION 1 ++++++++++

Why is $|E_A| = ({5 \choose 3} +{5 \choose 3} + {3 \choose 3} ){13 \choose 3}$

Rather than just $|E_A| = {5 \choose 3} +{5 \choose 3} + {3 \choose 3} ?$

So it seems my problem is in understanding exactly how to interpret an event. In my mind $|E_A|$ looks only at event E for A irrespective of what B draws, because if I want to calculate the probability of person A drawing only 3 balls of the same colour then surely the probability is independent of whatever balls B may draw??

I still don't get this.

The only way that I can somehow make sense of it is if we look at the solutions of the sample space, the elements of which is a solution set for all A and B selections, as the question states that the 2 events occur (have to and always will) one after the other.

My issue still is that working out the probability of $E_A$ will erroneously take into consideration the selections for B.

Ok, so it just occurred to me that in fact if we do work out the probability of $E_A$ we can do it in two ways 1) we only look at the draw of A and therefore work with a reduced sample space; and 2) when taking the entire sample space into consideration, thus:

For (1) we have:

$P(E_A) = \frac{| E_A |}{|S|}$ … where for S and thus E we look at the reduced sample space i.e. we don't take B into consideration.

$= \frac{{5 \choose 3} +{5 \choose 3} + {3 \choose 3}}{ {13 \choose 3} }$

$= \frac{21}{286}$

And for (2)

$P(E_A) = \frac{| E_A |}{|S|}$… where for S and thus E we look at the original sample space

$= \frac{ [ {5 \choose 3} +{5 \choose 3} + {3 \choose 3}] [{13 \choose 3}] } { {13 \choose 3} {13 \choose 3} }$

$= \frac{21}{286}$

So this seems to indicate once again that the only reason we take into consideration the draw of B is because the defined solution space enforces it?? I would think that there would be some more elegant way to express this in the Math as it is not evident when you just look at a variable defined as $E_A$? Surely I am missing something?

Let me try explaining this way:

Why is $|E_A| = [{5 \choose 3} +{5 \choose 3} + {3 \choose 3}] {13 \choose 3}$

Rather than just $|E_A| = {5 \choose 3} +{5 \choose 3} + {3 \choose 3} ?$

Consider for a moment that we have 5 Blue, 5 Red and 3 Green balls. I am trying to find the flaw in the following argument:

For $(E_A \bigcup E_B)^c (F_A \bigcup F_B)^c$

consider first $(E_A \bigcup E_B)^c = {E_A}^c \bigcap {E_B}^c$

Let's look at the elements in ${E_A}^c$:

$E_A(B, R, G) = { (1,1,1); (2,1,0); (2,0,1); (1,2,0); (0,2,1); (1,0,2); (0,1,2) }$ … And all permutations of the colours in every element

But ${E_B}^c$ has exactly the same set i.e.

$E_B(B, R, G) = { (1,1,1); (2,1,0); (2,0,1); (1,2,0); (0,2,1); (1,0,2); (0,1,2) }$

So that ${E_A}^c \bigcap {E_B}^c = { (1,1,1); (2,1,0); (2,0,1); (1,2,0); (0,2,1); (1,0,2); (0,1,2) }$ according to the enforcement of the AND operator.

Now for $(F_A \bigcup F_B)^c = {F_A}^c \bigcap {F_B}^c$

Let's look at the elements in ${F_A}^c$:

${F_A}^c = { (1,1,1); (3,0,0); (0,3,0); (0,0,3) }$ also

${F_B}^c = { (1,1,1); (3,0,0); (0,3,0); (0,0,3) }$

So that ${F_A}^c \bigcap {F_B}^ = { (1,1,1); (3,0,0); (0,3,0); (0,0,3) }$ …. And all permutations of the colours in every element.

Taking the above it follows that

$(E_A \bigcup E_B)^c (F_A \bigcup F_B)^c = ( { {E_A}^c \bigcap {E_B}^c } ) \bigcap ( { {F_A}^c \bigcap {F_B}^c } )$

$= { (1,1,1) }$ … as this is the only element that is in their intersection.

It should therefore follow that:

$| (E_A \bigcup E_B)^c (F_A \bigcup F_B)^c | = {5 \choose 1} {5 \choose 1} {3 \choose 1}$

BUT the solution in the memo is:

${5 \choose 1} {5 \choose 1} {3 \choose 1} . {5 \choose 1} {5 \choose 1} {3 \choose 1}$

If I can figure this out then most of my problems should be solved ... I hope ...

(b.i.1) A bag contains $13$ balls, of which $5$ are blue, $5$ are red, and $3$ are green. Person $A$ chooses three balls out of the bag, after which person $B$ does the same. If $A$ replaces the balls before $B$ chooses, find the number of ways $A$ can select three balls of the same colour.

The factor $$\binom{5}{3} + \binom{5}{3} + \binom{3}{3}$$ counts the number of ways person $A$ can select three blue balls or three red balls or three green balls. The factor $$\binom{13}{3}$$ counts the number of ways person $B$ can select three balls after $A$ has replaced the balls.

Hence, $$|E_A| = \left[\binom{5}{3} + \binom{5}{3} + \binom{3}{3}\right]\binom{13}{3}$$

(b.i.2) A bag contains $13$ balls, of which $5$ are blue, $5$ are red, and $3$ are green. Person $A$ chooses three balls out of the bag, after which person $B$ does the same. If $A$ replaces the balls before $B$ chooses, find the number of ways $A$ can select three balls of the same colour and $B$ chooses exactly two balls of the same colour.

The factor $$\binom{5}{3} + \binom{5}{3} + \binom{3}{3}$$ counts the number of ways that person $A$ can select three balls of the same color. The balls are then replaced. Hence, person $B$ is selecting $3$ of the $13$ balls in the bag. The factor $$\binom{5}{2}\binom{8}{1} + \binom{5}{2}\binom{8}{1} + \binom{3}{2}\binom{10}{1}$$ counts the number of ways person $B$ selects two blue balls and one of the other eight balls in the bag or two red balls and one of the other eight balls in the bag or two green balls and one of the other eight balls in the bag.

The choices are independent since person $A$ has placed the balls back in the bag before person $B$ chooses.

Hence, $$|E_A \cap F_B| = \binom{5}{3}\left[\binom{5}{2}\binom{8}{1} + \binom{5}{2}\binom{8}{1}+ \binom{3}{2}\binom{10}{1}\right]$$

(b.ii.2) A bag contains $13$ balls, of which $5$ are blue, $5$ are red, and $3$ are green. Person $A$ chooses three balls out of the bag, after which person $B$ does the same. If $A$ does not replace the balls before $B$ chooses, find the number of ways $A$ can select three balls of the same colour and $B$ chooses exactly two balls of the same colour.

If person $A$ selects three blue balls, then person $B$ selects three balls from a bag containing two blue balls, five red balls, and three green balls.

If person $A$ selects three red balls, then person $B$ selects three balls from a bag with five blue balls, two red balls, and three green balls.

If person $A$ selects three green balls, then person $B$ selects three balls from a bag with five blue balls and five red balls.

Since it is always possible for person $B$ to select two exactly balls of the same colour after person $A$ has removed three balls of the same colour from the bag, the events that person $A$ chooses three balls of the same colour and person $B$ chooses exactly two balls of the same colour are not disjoint. However, in this case, the selections that person $B$ can make depend on which balls person $A$ has selected. For instance, if person $A$ selects three green balls, it is no longer possible for person $B$ to select to select two green balls. However, it is possible for person $B$ to select two blue balls and one red ball or two red balls and one blue ball.

The stated answer is correct.

(a.i.5) A bag contains $13$ balls, of which $5$ are blue, $5$ are red, and $3$ are green. Person $A$ chooses three balls out of the bag, after which person $B$ does the same. If $A$ replaces the balls before $B$ chooses, find the number of ways $A$ can select exactly two balls of the same colour or $B$ makes a different selection than $A$.

The event $F_A$ means that person $A$ selects exactly two balls of the same color, regardless of whether person $B$ makes the same selection or not.

The event $G^C$ means that person $B$ does not make the same selection as person $A$, regardless of whether person $A$ has selected three balls of the same color.

Hence, a selection is in $F_A \cup G^C$ if

• person $A$ selects exactly two balls of the same colour and person $B$ makes exactly the same selection: $F_A \cap G$
• person $A$ selects exactly two balls of the same colour and person $B$ makes a different selection: $F_A \cap G^C$
• person $A$ does not select two balls of the same colour and person $B$ makes a different selection: $F_A^C \cap G^C$

What is not in the union is a selection in which person $A$ does not select exactly two balls of the same colour and person $B$ makes exactly the same selection as $A$: $F_A^C \cap G$. The number of outcomes of this event must be subtracted from the $$\binom{13}{3}\binom{13}{3}$$ possible ways for $A$ and $B$ to each select $3$ of the $13$ balls in the bag, so the stated answer is incorrect.

Since there are $$\binom{5}{2}\binom{8}{1} + \binom{5}{2}\binom{8}{1} + \binom{3}{2}\binom{10}{1}$$ ways for person $A$ to select exactly two balls of the same colour and $\binom{13}{3}$ ways for person $A$ to select three balls, there are $$\binom{13}{3} - \left[\binom{5}{2}\binom{8}{1} + \binom{5}{2}\binom{8}{1} + \binom{3}{2}\binom{10}{1}\right]$$ ways for person $A$ to not select exactly two balls of the same colour. For each such selection, there is exactly one way for person $B$ to make exactly the same selection. Hence, $$|F_A \cup G^C| = \binom{13}{3}\binom{13}{3} - \binom{13}{3} + \binom{5}{2}\binom{8}{1} + \binom{5}{2}\binom{8}{1} + \binom{3}{2}\binom{10}{1}$$

• Thanks for the response. I have a couple of questions especially for your last solution, but want to address another issue first. Please see the addition I added to the original question. Sep 14, 2017 at 14:32
• The statement "Person $A$ chooses randomly three balls out of the bag. Afterwards, Person $B$ does the same" means that the sample space consists of all selections of three balls by person $A$ and three balls by person $B$. That is why we count the number of selections person $B$ makes when we calculate $|E_A|$. As you point out, if the question asked for the probability of $E_A$, the number of selections person $B$ makes would not matter since it would cancel out. However, we are asked for the number of occurrences of the event rather than the probability. Sep 14, 2017 at 15:08
• Right thanks. So how would that affect my argument in the addition I added? Is it correct to say: ${E_A}^c = { (1,1,1); (2,1,0); (2,0,1); (1,2,0); (0,2,1); (1,0,2); (0,1,2) }$ And, ${E_B}^c = { (1,1,1); (2,1,0); (2,0,1); (1,2,0); (0,2,1); (1,0,2); (0,1,2) }$ Therefore, ${E_A}^c \bigcap {E_B}^c = { (1,1,1); (2,1,0); (2,0,1); (1,2,0); (0,2,1); (1,0,2); (0,1,2) }$ Sep 14, 2017 at 18:18
• So that, $| {E_A}^c \bigcap {E_B}^c | = {5 \choose 1} {5 \choose 1} {3 \choose 1 } + {5 \choose 2} {5 \choose 1} + {5 \choose 2} {3 \choose 1} + {5 \choose 1} {5 \choose 2} + {5 \choose 2} {5 \choose 1} + {5 \choose 1} {3 \choose 2} + {5 \choose 1} {3 \choose 2}$ So there would be no multiplication with ${13 \choose 3}$ because both A and B were considered in the top equation? Sep 14, 2017 at 18:19