Probability Question: 3 red balls, 4 green balls, and 4 blue balls are lined up in a random order.

Question

I have been trying to do a Probability exercise. I was able to solve first 2-3 points of the exercise but stuck on the conditional exercise. Please help with little explanation.

Three red balls, four green balls, and four blue balls are lined up in a random order (that is, each ordering is equally likely).

1. How many possible ways are there to order these balls, if we can distinguish all balls? (Assume for instance that each ball carries a number between 1 and 11.)
2. What is the probability that in a random line-up all the balls of each color are together, that is, first there are all the balls of one colour, then the ones of another colour, and finally all the ones of the third colour?
3. What is the probability that the last three balls are red?
4. Knowing that the third ball is green, what is the probability that the last two balls are red
5. Knowing that the ninth ball is blue, what is the probability that the last three balls are blue?
6. Knowing that the ninth ball is green, what is the probability that the last three balls are blue?

1. 11!

2. (3! 3! 4! 4!) / 11!

3. (3! 8!) / 11!

4. I am stuck... I have tried different combinations but not able to reach expected conclusion.

What I have tried is this:

Let A be the last two balls are red. Let B be the 3rd ball is green. Find P(A|B).

To Find A: 9!2!/11! = 2/110

To Find B: 4C1*8!/11! = 4/990

Then apply the P(A|B) = P(A ∩ B) / P(B) ==> after plugging ==> 2/110. Which is same as P(A). That cannot be. I don't know I think, I am plugging the formula wrong... because whatever the value I assign to P(B) is gonna cancel with denominator and nominator.

Later I have tried to find P(A) and P(B) in other ways but...

• attempts to find A:
  • 3!6!2!/11! = 1/4200
  • (3C2 * 2 * 8!) / 11! = 1/165
• attempts to find B:
  • (4C1 * 3 * 10!) / 11!

Then I have tried useless different combination ==> P(A∩B)/P(B)

Later I release these are dependent events so I have to use 1 - P(A) + p(B). But I didn't get the release.

Please tell me also what I have done wrong? What should I set my mindset for these kinds of problems?

Can you please solve the 4th point of the exercise with explanations?

Answer 1

If you fix the one green ball at the third position there are 10 balls left to arrange on any position 1 2 4 ... 11. But we can pretend we just arrange the 10 balls and insert the green one afterwards. So this should be the same as asking the probability for the two last balls red in the setting 3 red balls, 3 green balls and 4 blue balls.

Edit: Exact answer:

Assuming the different colors are not separable and using multinomial coefficients.

My first approach results in $\binom{8}{1,3,4}/\binom{10}{3,3,4}=\frac{1}{15}$. This is the number of arrangements where the two red balls at the end are fixed and the green one on the third position divided by the number of arrangements where only the green ball is fixed.

With Bayes rule:

Number of total arrangements $N=\binom{11}{3,4,4}$

$P(B)$, where $B$ is the event, that there is a green ball on the third position: $\frac{\binom{10}{3,3,4}}{N}$

$P(A\cap B)$, where $A\cap B$ is the fixed green ball and the two fixed red ones, $\frac{\binom{8}{1,3,4}}{N}$.

$P(A|B)=\frac{P(A\cap B)}{P(B)} = \frac{1}{15}$