Find probability that at least one box contains a total of $6$ or more balls when $10$ balls are selected

Question

I have a total of 50 balls numbered 1 to 50. I have 4 boxes:

• Box 1 range 1 to 15
• Box 2 range 16 to 30
• Box 3 range 31 to 45
• Box 4 range 46 to 50

I pick 10 balls at random from those numbered 1 to 50 each round. I place each ball in one of the 4 boxes according to range.

e.g

If I pick the number 13 ball, place it in Box 1 If I pick the number 30 ball, place it in Box 2

What is the probability that Box 1 or Box 2 or Box 3 contain a total of 6 or more balls?

Please explain briefly with the equation.

Thanks.

Answer 1

The context shows, that the picking is without repetition, so:

1. Notice, that if we pick 10 balls, there can be at last one box with 6 or more bals in it. Events where there are 6 or more balls in box n are disjoint, so $$P=P(6 \text{ or more in Box1 or Box2 or Box3}) = \\ P(6 \text{ or more in Box1})+P(6 \text{ or more in Box2})+P(6 \text{ or more in Box3})$$
2. Events for box1, box2 and box3 are the same - 15 balls are in the range, 35 bals are outside the range. So $$P(6 \text{ or more in Box1})=P(6 \text{ or more in Box2})=P(6 \text{ or more in Box3}) = Q$$
3. The space of all pickings have $$|\Omega|=\binom{50}{10}$$ elements
4. For $k\geq 6$ event where there are exactly $k$ balls in box n can be described as: Pick $k$ balls of 15 that fits in box n and $10-k$ balls of 35 that fits in the other boxes. It can be done in $$a_k=\binom{15}{k}\binom{35}{10-k}$$ ways.
5. Obviously the events, where there are exactly 6,7,8,9 and 10 elements in box n are disjoint, so $$Q=\frac{\sum_{k=6}^{10}a_k}{|\Omega|}$$
6. Finally $$P=3Q = \frac{3\sum_{k=6}^{10}\binom{15}{k}\binom{35}{10-k}}{\binom{50}{10}}$$