6 Red balls, 7 Blue balls, ways to choose 5 balls so there are more red balls chosen.

Question

My thoughts are the following:

P(Ways to choose 5 balls so there are more red balls) = P(3 red are chosen) + P(4 red are chosen) + P(5 red are chosen).

My intuition says to do this problem considering each subproblem with conditional probability (eg, P(3 are chosen) = (6/13)(5/12)(4/11)(7/10)(6/9)).

I feel like there is an easier combinatorial answer, but I haven't dealt with this type of problem in a few months.

Answer 1

The total number of ways to choose $5$ balls is $\binom{6+7}{5}$.

If more red balls are chosen (than blue balls), then exactly one of the following must apply:

• $5$ red balls were chosen. There are $\binom{6}{5}$ possibilities for this case.
• $4$ red balls and $1$ blue ball were chosen. There are $\binom{6}{4}\cdot\binom{7}{1}$ possibilities for this case.
• $3$ red balls and $2$ blue balls were chosen. There are $\binom63\cdot\binom72$ possibilities for this case.

Hence, the probability is given by

$$\frac{\binom65+\binom64\cdot\binom71+\binom63\cdot\binom72}{\binom{13}5}=\frac{59}{143}\simeq41.26\%$$