1. ## Confused about basic probability

A box contains 5 white balls, 3 red balls and 2 green balls.

The balls are taken from box. With replacement and taking a total of 3 balls,

P(all balls have different colour)=(5/10)x(3/10)x(2/10) x 3! =0.18

This answer is the same as at the back of the book. My explanation is that a white, red or green ball must be taken and there are 3! ways of arranging the 3 scenarios.

Is this correct????

Then secondly,
Without replacement and taking a total of 5 balls

My wrong method:
P(exactly 2 balls are white) = (5/10)x(4/9)x(5/8)x(4/7)x(3/6) x 5!

This probability cannot be correct since it is above 1. What is the loophole that I did not see???

I dont need the answer to this question (I can solve using counting total number of cases) but I want to know the loophole in my thinking.

Thanks,

2. Hello, qazxsw11111!

A box contains 5 white balls, 3 red balls and 2 green balls.

Three balls are taken from box with replacement.

My work:

$\displaystyle P(\text{different colors}) \:=\:\frac{5}{10}\times \frac{3}{10} \times \frac{2}{10} \times 3! \:=\:0.18$

My explanation is that a white, red or green ball must be taken
and there are 3! ways of arranging the 3 scenarios.

Correct!

Five balls are taken without replacement.

My wrong method:

$\displaystyle P(\text{exactly 2 White} \:=\: \frac{5}{10} \times\frac{4}{9} \times\frac{5}{8}\times\frac{4}{7} \times \frac{3}{6} \times {\color{red}5!}$

This probability cannot be correct since it is above 1.
What is the loophole that I did not see?

We want 2 Whites and 3 Others.

There are: .$\displaystyle _5C_2 \:=\:\frac{5!}{2!\,3!} \:=\: 10$ scenarios ... not 120.

3. Thank you for your answer. But can you help to clarify when to add factorial sign and when not to?

Thanks.