Math Is Fun Forum

NESIC

Member

Registered: 2010-09-25

Posts: 16

Probability

Dear All kindly Guide me if solution is wrong

A bag contains 14 identical balls, 4 of which are red, 5 black and 5 white. 6 balls are
drawn from the bag. Find the probability that,
a. Three are red.
b. At least 2 are white.

Solution:
A.
Total no of possible outcomes 14!/6!8!=3003
n(A)=480
P(3red)=480/3003=.159

For Part B Pls guide me how i start

Last edited by NESIC (2010-10-07 06:15:06)

Bob

Administrator

Registered: 2010-06-20

Posts: 10,583

Re: Probability

Hi Nesic

I've changed this post in the last 30 minutes so if you read it just now, please disregard.  Here's my edited version.

Here is my method:

Let us look first at how to pick one red, then another red, then another red, then a non-red, then another non-red then another non red.

P(Red1) = 4/14        P(red2) = 3/13         P(red3)  = 2/12        P(non-red1) = 10/11        P(non-red2) = 9/10      P(non-red3) = 8/9

so P(RRRNNN) =

But the reds and non-reds could come in a different order.  Let us look at one more.

P(RNRNRN) =

As you can see this is the same calculation again.  So we do not need to work out all the different ways of getting 3 Rs and 3Ns.

Just do it once and multiply by the number of ways of shuffling the Rs and Ns about.

There are

of these.

So final answer =

I should have calculated this first time because this is exactly your answer!

To do the second part, use this method to calculate P(no whites) and P(1 white).  Add together and subtract from 1.

Hope that helps,

Bob

Last edited by Bob (2010-10-07 09:10:09)

---

Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob

bobbym

bumpkin

From: Bumpkinland

Registered: 2009-04-12

Posts: 109,606

Re: Probability

Hi All;

Part A is correct by the hypergeometric distribution. Now for the tough one.

For B I am getting:

P(at least 5 white) = P(2 white) + P(3 white) +P(4 white) + P(5 white)

P(at least 5 white) = 60 / 143  + 40 / 143 + 60 / 1001 + 3 / 1001 = 109 / 143

We use the formula for the hypergeometric distribution for the above. For wanted = 2 to 5.

---

In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

soroban

Member

Registered: 2007-03-09

Posts: 452

Re: Probability

. .

. .

.

NESIC

Member

Registered: 2010-09-25

Posts: 16

Re: Probability

thanks soroban and all last one clear all things