probability

**allrighty** · January 27th 2008, 01:41 PM

There are 5 different colors of balls: white, black, blue, red, green. We randomly pick 6 balls. Each ball has a probability of 0.2 of getting each of the 5 colors. What is the probability that, from the 6 balls picked, there are white balls and black balls?

**Jhevon** · January 27th 2008, 01:56 PM

Originally Posted by allrighty

There are 5 different colors of balls: white, black, blue, red, green. We randomly pick 6 balls. Each ball has a probability of 0.2 of getting each of the 5 colors. What is the probability that, from the 6 balls picked, there are white balls and black balls?

find the probabilities for each case and add them up

by (all possible) cases, i mean:

probability of choosing 1 white, 0 black, 1 blue, 1 red, 3 green

probability of choosing 1 white, 0 black, 0 blue, 2 red, 3 green

probability of choosing 1 white, 0 black, 2 blue, 0 red, 3 green

.
.
.
.

or we could do:

1 - probability of choosing no white and no black ball

so all you have to worry about are the number of ways you can choose 6 balls among the three remaining colors. there is a formula for that sort of thing. can't remember right now, i'll have to look it up. but when you do get it, just multiply the answer by 0.2 and that will give you the probability of choosing no white and no black ball

**a tutor** · January 27th 2008, 02:15 PM

Originally Posted by Jhevon

so all you have to worry about are the number of ways you can choose 6 balls among the three remaining colors

Is that right? A white AND a black are required.

I have another method but it's messy.

You could have for example 2W, 1B and 3 others 6!/(3!2!) ways each with probability 0.2^3 0.6^3

or 5W, 1B 6!/5! ways with probability 0.2^6

Having said all that, I'm a bit rubbish at probability so I'll wait and see what people say.

**Jhevon** · January 27th 2008, 02:25 PM

Originally Posted by a tutor

Is that right? A white AND a black are required.

yes, i believe so. what i said was i want to find the probability of having 0 white AND 0 black. and then take 1 minus that probability to find the probability of at least 1 white or 1 black. i think what i described does the trick... but then again, i'm a noob when it comes to probability as well

**a tutor** · January 27th 2008, 02:37 PM

Original question said..

Originally Posted by allrighty

What is the probability that, from the 6 balls picked, there are white balls and black balls?

and you said..

Originally Posted by Jhevon

to find the probability of at least 1 white or 1 black.

**Jhevon** · January 27th 2008, 02:39 PM

Originally Posted by a tutor

Original question said..

and you said..

ah yes. my bad. i didn't see the "and." i was under the impression if we have either or we were good...

**Soroban** · January 27th 2008, 05:14 PM

Hello, allrighty!

I think I've solved it . . .

There are 5 different colors of balls: white, black, blue, red, green.
We randomly pick 6 balls.
Each ball has a probability of 0.2 of getting each of the 5 colors.
What is the probability that, from the 6 balls picked, there are white balls and black balls?

The opposite of "some White and some Black" is "no White or no Black".

To get no White, we must pick six balls from the other four colors.
. . Then: . $P(\text{0 White}) \:=\:(0.8)^6$

To get no Black, we must pick six balls from the other four colors.
. . Then: . $P(\text{0 Black}) \:=\:(0.8)^6$

To get no White and no Black, we pick six balls from the other 3 colors.
. . Then: . $P(\text{0 White} \,\wedge \,\text{0 Black}) \:=\;(0.6)^6$

Hence: . $P(\text{0 White} \,\vee \,\text{0 Black}) \;=\;P(\text{0 White}) + P(\text{0 Black}) - P(\text{0 White} \,\wedge \,\text{0 Black})$
. . . . . . . . . . . . . . . . . . . . $= \quad\;\;(0.8)^6\quad \;+ \;\quad(0.8)^6 \qquad- \qquad(0.6)^6$
. . . . . . . . . . . . . . . . . . . . $= \qquad 0.4777632$

Therefore: . $P(\text{some White and some Black}) \;=\;1 - 0.4777632 \;=\;\boxed{\:0.522368\:}$

**a tutor** · January 28th 2008, 02:11 AM

A neat solution Soroban.

I got the same answer rather more clumsily.

I wrote a quick easy program to do it the way I mentioned above.

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