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

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

Step-by-step 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.