# Probability: 10 balls in vessel (5 white, 5 black); find prob of picking 3 w/o replacement & not all same color

##### Junior Member
I may miss something that I must understand it in probability, so I attach a question in the same context and please tell me if I'm correct or not .. we have socks in a vessel : 5 red, 4 black, 3 red, 2 white, 1 orange.
what's the minimal number of socks that we must take out to assure 100% that we have two socks with different colors?

my answer is ; 5 red to take out and then one sock of other brands so we have in total

6 socks we must get out .. right?!

#### pka

##### Elite Member
I've the same problem as previous thread but it's a lil different and please need a help to solve it.
there're 10 balls in the vessel which 5 from them are white and the other 5 are black, what's the probability of 3 rounds of taking out balls (without returning the ball) to not getting out balls with the same color ? (in other words we musn't get three balls with the same color)
The easy way to do this is to find the complement. Let W be all white & B be all black.
$$\displaystyle \mathcal{P}(B)=\frac{5}{10}\cdot\frac{4}{9}\cdot\frac{3}{8}$$. The same for all white. Because those are disjoint events adding then gives you the probability of either. Now find the complement.

#### Dr.Peterson

##### Elite Member
I may miss something that I must understand it in probability, so I attach a question in the same context and please tell me if I'm correct or not .. we have socks in a vessel : 5 red, 4 black, 3 red, 2 white, 1 orange.
what's the minimal number of socks that we must take out to assure 100% that we have two socks with different colors?

my answer is ; 5 red to take out and then one sock of other brands so we have in total

6 socks we must get out .. right?!
This actually is not a probability problem at all! Probability can't give you certainty. Rather, this is a "pigeonhole principle" question.

But you are correct. If you took out only 5 socks, they might all be the same color, but taking 6 ensures that there are at least two different colors among them.

#### JeffM

##### Elite Member
1/6 + 1/6 + 1/6
But 1/6 is not the probability of anything at all relevant. What do you think has a probability of 1/6?

The most intuitive way to do this problem is to realize that there are 8 possible NON-overlapping cases, namely

WWW
BWW
WBW
WWB
WBB
BWB
BBW
BBB

Are those cases equiprobable? No.

What is the probability of the first case?

$$\displaystyle \dfrac{5}{10} * \dfrac{4}{9} * \dfrac{3}{8} = \dfrac{60}{720} = \dfrac{1}{12}.$$

Do you follow that?

What is the probability of the second case?

$$\displaystyle \dfrac{5}{10} * \dfrac{5}{9} * \dfrac{4}{8} = \dfrac{100}{720} = \dfrac{5}{36}.$$

Do you follow that?

Can you work out the probabilities of the remaing 6 cases?

Of the 8 cases, which ones give you a result with more than one color?

So what is the probability of more than one color?

This approach is NOT the most efficient way to get the answer, but it may explain the answer.

#### Jomo

##### Elite Member
5 white and 5 black.

You want to find P (drawing 3 balls w/o replacement and they are NOT all the same color)
First I would find P (drawing 3 balls w/o replacement and they are all the same color) = P(3 white or 3 black) = P(3 white) + P(3 black) = 2*P(3 white)

Now P(3 white) = (5/10)(4/9)(3/8) = 1/12. So P(3 white or 3 black) =1/6

Then what would P (drawing 3 balls w/o replacement and they are NOT all the same color) = ?

#### Ryan\$

##### Junior Member
thanks you all I understand every thing in that subject!
for the last comment, I should use compliment (1- prevented event)

• Jomo

#### pka

##### Elite Member
thanks you all I understand every thing in that subject!
for the last comment, I should use compliment (1- prevented event)
YES $$\displaystyle 1-\mathcal{P}(W\cup B)$$ i.e. one minus probability of all white or all black.