I would really appreciate some help in the following probability question. I translated it so I'm sorry in advance for English mistakes.

There are $6$ children in a family and many candies in $4$ colors - blue, red, green, pink. The $2$ eldest children get to choose $2$ candies each. The $4$ other children get to choose $1$ candy apiece. (The same candy type can be chosen by a few children).

a. Count the number of choices.

b. What is the probability that exactly $2$ children choose blue candies?

My attempt at solution:

a. $$\binom{4}{2}^2 \cdot 4^4 = 6^2 \cdot 4^4$$

This solution is correct according to the solutions given to me.

b. I tried to solve it in the following way:

Dividing the question into 3 cases:

$(1)$ The number of cases in which the two eldest children choose blue candies: $$3^4$$

$(2)$ The number of cases in which two younger children choose blue candies: $$3^2 \cdot 3^2 = 3^4$$

$(3)$ The number of cases in which one blue candy is chosen by an older child and one by a younger: $$2 \cdot 4 \cdot 3 \cdot 3^3 = 2 \cdot 4 \cdot 3^4$$

I sum all the case to get $$10 \cdot 3^4$$

The final solution I get is: $$\frac{10 \cdot 3^4}{6^2 \cdot 4^4} = 0.088$$

My issues with this question is that in the solutions given to me, the number of options in case $(1)$ is calculated as: $$3^2 \cdot 3^4$$ and in case $(2)$: $$3^2 \cdot \binom{4}{2} \cdot 3^2$$

My question is - why is this solution correct?

Any help would be greatly appreciated.

Thank you in advance!!


In the case 2 you must also consider the number of ways in which we can select 2 kids from the younger ones that is:

$$4\choose 2$$


  • $\begingroup$ Oh sorry I missed that, thank you! $\endgroup$ – PhysicsPrincess Mar 31 '19 at 18:27

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