×

If in a box there are 3 blue balls, 4 red balls, 7 yellow balls and 2 green balls, then how many times we have to take a ball at random so that we would get 2 balls of the same color?

Note by Dina Andini Sri Hardina
4 years, 7 months ago

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold
- bulleted- list
• bulleted
• list
1. numbered2. list
1. numbered
2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in $$...$$ or $...$ to ensure proper formatting.
2 \times 3 $$2 \times 3$$
2^{34} $$2^{34}$$
a_{i-1} $$a_{i-1}$$
\frac{2}{3} $$\frac{2}{3}$$
\sqrt{2} $$\sqrt{2}$$
\sum_{i=1}^3 $$\sum_{i=1}^3$$
\sin \theta $$\sin \theta$$
\boxed{123} $$\boxed{123}$$

Sort by:

I think your question needs editing, it should rather be "how many minimum no. of balls one needs to pick to assure that he gets atleast 2 balls of the same colour". Then the answer would be 5 using PHP.

- 4 years, 7 months ago

i don't know, i just write the original question. but thanks!:)

- 4 years, 7 months ago

If the problem is on expectation, I get $$\dfrac{212}{49}$$ balls to be picked on average.

- 4 years, 6 months ago

Sorry, my above answer is a blunder! I find the expectation to be $$\frac{8741}{2730}$$

- 4 years, 6 months ago

5 . Because if we need 4 chances to pick a ball of every colour then another chance will surely be the second ball of any colour.

- 4 years, 7 months ago

Thanks:D

- 4 years, 7 months ago