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A bowl contains 20 red balls and 20 blue balls. A man selects balls at random without looking at them. How many balls must he select to be sure of having at least five balls of the same color

User Kerrine
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1 Answer

4 votes

Answer:

we will select 9 balls to be sure of having at least five balls of the same color.

Explanation:

Given that:

The bowl contains (20 red + 20 blue) balls = 40 balls

If we are to select 5 balls out of the 40 balls.

Let R = red and B = blue

The combinations can be:

5R + 0 B

4R + 1 B

3R + 2 B

1R + 4 B

0 R + 5 B

From above, we will notice that there are valid proof that there are 5 balls of the same color.

If we select six balls, we have:

= (6R + 0 B ), (5R + 1 B ), (4R + 2 B ), (3R + 3 B ), (2R + 4 B ), (1R + 5 B ), (0R + 6 B )

If we select seven balls, we have:

= (7R + 0 B ), (6R + 1 B ), (5R + 2 B ), (4R + 3 B ), (3R + 4 B ), (2R + 5 B ), (1R + 6 B ), (0R + 6 B ).

If we select eight balls, we have:

= (8R + 0 B ), (7R + 1 B ), (6R + 2 B ), (5R + 3 B ), (4R + 4 B ), (3R + 5 B ), (2R + 6 B ), (1R + 7 B ), (0R + 8B)

Now, selecting nine balls, we have the following combinations:

= (9R + 0 B ), (8R + 1 B ), (7R + 2 B ), (6R + 3 B ), (5R + 4 B ), (4R + 5 B ), (3R + 6 B ), (2R + 7 B ), (1R + 8B), (0R + 9B).

Thus, we can see that all the combinations comprise of 5 balls with at least one color.

Therefore, we will select 9 balls to be sure of having at least five balls of the same color.

User MaximeF
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