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A bag contains 2 white b***s,3 black b***s and 4 red b***s. In how many ways can 3 b***s be drawn form the bag, if at least one black ball is to be included?

Is this answer correct? (6 2)(3 1)+(6 1)(3 2)+(6 0)(3 3)

ultmage11234  Jan 30, 2015

Best Answer 

 #4
avatar+330 
+18

 

$$\Text {Solution}\\\

\Text {Total number of ways to choose 3 b***s from 9}\\
\Text {nCr(9,3) = 84 (where the order is irrelevant).}\\

\Text {Six non-black b***s and three black b***s.} \\


\text {nCr(6,2)*ncr(3,1) = 45}\\
\text {Number of ways to select 2 of 6 non-black b***s and select 1 of 3 black b***s.} \\

\text nCr(6,1)*ncr(3,2) = 18 } \\
\hspace{5ex}\; \text {Number of ways to select 1 of 6 non-black b***s and select 2 of 3 black b***s.}\\

\text {nCr(6,0)*ncr(3,3) =1 } \\
\text {Number of ways to select 0 of 6 non-black b***s and select 3 of 3 black b***s.}$$

 

(45+18+1) = 64 Total number of ways to choose 1, 2 or 3 black b***s of 3 b***s from 9 b***s where 3 are black and 6 are non-black.

The question asks for Permutation, but this is really a Commutation problem not a Permutation problem. The Binomial distribution solution should apply.

 

If these b***s were unique such as with numbers along with the colors, then Alan’s solution would apply. This means the drawing order of the black b***s along with the order of the non-black b***s is unique in each instance, and there would be many more arrangements.

 

~DQD~

DavidQD  Jan 30, 2015
 #1
avatar+7188 
+5

Hmmmm...I'm not a great one with probability,  to be honest

happy7  Jan 30, 2015
 #2
avatar+26750 
+10

There are 3 black b***s and 6 non-black b***s.

 

First think of how many ways there are to draw three b***s (without replacement, presumably) such that you get no black b***s.

There are 6 ways to draw the first. For each of these there are 5 ways to draw the second. For each of these pairs there are 4 ways to draw the third.  Thus there are 6*5*4 = 120 ways to draw three b***s such that none are black.

 

In total there are 9*8*7 =  504 ways to draw three b***s from the bag, irrespective of colour, so there must be 504 - 120 = 384 ways to draw three b***s from the bag such that at least one is black.

.

Alan  Jan 30, 2015
 #3
avatar+52 
+5

So this formula does not work?

nCr

C = combination

n = total number of ojects that you can choose from

r = the number that you need to choose

 1blackball and 2 nonblack b***s + 2blackballs and 1 nonblack ball + 3blackballs and 0 nonblack b***s

(3C1)(6C2)+(3C2)(6C1)+(3C3)(6C0)

ultmage11234  Jan 30, 2015
 #4
avatar+330 
+18
Best Answer

 

$$\Text {Solution}\\\

\Text {Total number of ways to choose 3 b***s from 9}\\
\Text {nCr(9,3) = 84 (where the order is irrelevant).}\\

\Text {Six non-black b***s and three black b***s.} \\


\text {nCr(6,2)*ncr(3,1) = 45}\\
\text {Number of ways to select 2 of 6 non-black b***s and select 1 of 3 black b***s.} \\

\text nCr(6,1)*ncr(3,2) = 18 } \\
\hspace{5ex}\; \text {Number of ways to select 1 of 6 non-black b***s and select 2 of 3 black b***s.}\\

\text {nCr(6,0)*ncr(3,3) =1 } \\
\text {Number of ways to select 0 of 6 non-black b***s and select 3 of 3 black b***s.}$$

 

(45+18+1) = 64 Total number of ways to choose 1, 2 or 3 black b***s of 3 b***s from 9 b***s where 3 are black and 6 are non-black.

The question asks for Permutation, but this is really a Commutation problem not a Permutation problem. The Binomial distribution solution should apply.

 

If these b***s were unique such as with numbers along with the colors, then Alan’s solution would apply. This means the drawing order of the black b***s along with the order of the non-black b***s is unique in each instance, and there would be many more arrangements.

 

~DQD~

DavidQD  Jan 30, 2015
 #5
avatar+87301 
+10

Here's my take on this one......

We're just counting sets of things, and order witin sets doesn't matter, so this is a combination problem, rather than a permute problem.

The number of possible sets made by choosing any 3 b***s from 9 =C(9,3) = 84

And the number of possible sets made by choosing no black b***s is just C(6,3) = 20

So, the number of possible sets made by choosing at least one black ball = 84 - 60 = 64

 

{I'm not great at probability, so I really don't know if i'm correct, ot not....!!!}

 

CPhill  Jan 30, 2015
 #6
avatar+26750 
+10

I agree, my answer assumed each ball was separately identifiable, whatever the colour.

Alan  Jan 30, 2015
 #7
avatar+26750 
+15

As a slight tweak to the problem, suppose the question was "what is the probability that at least one black ball is drawn in three b***s?"

 

Prob that 1st ball is non-black = 6/9

Prob that next ball is non-black = 5/8

Prob that final ball is non-black = 4/7

So probability of at least 1 black ball = 1 - 6*5*4/(9*8*7) = 16/21

 

Using DavidQD and Chris's values above:  Probability of at least 1 black ball = 64/84 = 16/21

 

Using my answer to the original problem:  Probability of at least 1 black ball = 384/504 = 16/21

.

Alan  Jan 31, 2015
 #8
avatar+92777 
+10

This is my 2 cent worth 

There are a number of was to look at this question.  If it was part of a probablitiy question you would assume every ball was numbered as well as coloured

BUT this is NOT a probablily question so I am going to interprete it differently

Added after:  The title is Permutation which means that Order Does Count

-----------------------------------

First interpretation:  order does not count.  Here are my choices.    

BBB, BBR, BBW, BRR, BWW, BRW         That makes 6 outcomes

--------------------------------

Now I am going to look at if order does count.

Here are the choices

BBB

BBR or BBW

BRB or BWB

RBB or WBB

Now if I  chose just 2 non-black ones then I can have  WW,WR,RW, or RR  now the black one can be 1st second or third      so that is   4*3=12

Total = 1+6+12=19 ways       [This is the only one with any validity out of my 3]

--------------------------------------------------------------

NOW LETS LOOK AT IT FROM AN ORDER DOES NOT COUNT PROBABILITY PERSPECIVE

Total number of ways that you can choose 3 b***s from 9 is 9C3=84

Total number of ways of not choosing any black ones is   6C3 = 20

There for total number of ways of haveing a black one in the mix must be 84-20=64

(This is the same answer as most the others got I think)

 

Lets do this the long way and see if there is any agreement

all black   3C3=1

2black and 1 other = 3C2*6C1=3*6=18

1black and 2 other = 3C1*6C2=3*15=45

1+18+45=64                                      GOOD IT IS THE SAME

P(choosing at least 1 black out of 3 where order doesnt count.) = 64/84

 

64 IS NOT THE CORRECT ANSWER BECAUSE THE TITLE OF THIS QUESTION IS PERMUTATION.

64 IS THE ANSWER TO A COMBINATION QUESTION SO IT IS NOT CORRECT  !!

---------------------------------------------------------------------

 

Now how about if order does count from a proabability perspective

It is all getting a bit much now.     

---------------------------------------------------------------------------

Melody  Jan 31, 2015
 #9
avatar+92777 
+10

Alan I had an answer similar to your first one but I got in a mess and scrubbed it.

If you are looking at this from a probabliliy perspective  I think that (or something very similar ) would be correct

but

from the choosers point of view without worrying about probabilities there are 19 possible outcomes.

 

So Ultmage  your answer is not correct.    

Melody  Jan 31, 2015
 #10
avatar+87301 
+10

ultmage11234's answer is correct....if we consider this to be a combination problem...

If we consider it to be a permute....we have

P(9,3) - P(6,3) = 384 ways

This is a little more difficult to see....

Consider the set of permutes that include one black ball...

If we let it occupy the first position in the set, we have 6 other b***s to choose from in the second position and 5 more to choose from in the 3rd position. Thus, 3 x 6 x 5 = 90

And since a black ball can occupy any one of three positions, this gives 270 ways to permute a set which includes one black ball.

For a set that includes two black b***s, we have 3 ways to choose one for the first position and two ways to choose one for the second position and 6 ways to choose a final ball. But the non-black ball can occupy any one of three positions. So the total number of permutes in a set containing 2 black b***s = 3 x 2 x 6 x 3 = 108

And we have 6 ways to arrange 3 black b***s in a set that just contains 3 black b***s.

So...the total number of permutes of all the sets containing at least one black ball is just

270 + 108 + 6  = 384   .....just as we speculated!!!

{Constructive criticism is welcome.....since I am not a probability "expert" }

 

CPhill  Jan 31, 2015
 #11
avatar+92777 
+10

The title of the question is PERMUTATION.

 

So order DOES count    

 

With permutation questions the b***s are usually marked with invisable numbers so that the 3 black b***s can be thought of seperately.  This is necessary if you are going to compute probablities from it.  This is what chris and Alan attempted to do  (their answers may well be correct - I have not thought that through yet)

 

HOWEVER if you are only interested in the outcomes and not the probabilities attacked to them I am still standing by my answer of 19.

 

The question asker could have considered this either way BUT order definitely counts.  Quite likely it was not my answer of 19 that they were looking for.

 

Chris did you look at the working for my answer of 19.  I simply looked at the possibilities and added them together.  My logic is very straight forward.  

Melody  Jan 31, 2015
 #12
avatar+87301 
+5

Yep, Melody...your answer is correct if we consider that the b***s are indistinguishable....

Alan and I have taken the "numbered" ball approach in considering permutes.....

Which is "correct??"

...Depends upon your assumption.....

 

CPhill  Jan 31, 2015
 #13
avatar+52 
0

I mightve confused people with the title, it might not necessarily be a premutation question in regards to the problem. My assumption was that it is a premutation question but im not entirely sure. It could be either premutation or combination i guess. Which one does it come closer to? Try to solve it disregarding the title!

ultmage11234  Jan 31, 2015
 #14
avatar+87301 
0

Well...if it was a combination....it would cut down on the controversy, here....LOL!!!

 

 

CPhill  Jan 31, 2015
 #15
avatar+52 
+5

Yeah for sure, atleast i learned a few thing like distinguishing probalitity problems like these xD

ultmage11234  Jan 31, 2015
 #16
avatar+52 
+5

THANKS EVERYBODY!!

ultmage11234  Jan 31, 2015
 #17
avatar+92777 
+5

I don't think that we have finished with this question yet ultmage   

Melody  Jan 31, 2015
 #18
avatar+87301 
+5

OK.....at least your question generated a lot of interest...!!

 

CPhill  Jan 31, 2015
 #19
avatar+52 
+5

So ultimately is my answer right? Many of you suggest the question to be a combination type of problem.

 1blackball and 2 nonblack b***s + 2blackballs and 1 nonblack ball + 3blackballs and 0 nonblack b***s

(3C1)(6C2)+(3C2)(6C1)+(3C3)(6C0) = 64

ultmage11234  Jan 31, 2015
 #20
avatar+92777 
+5

 

Hi Ultmage11234,

welcome to the web 2 forum :)

 

Oh if you are not sure of the wording then there are 4 different interpretations.  All equally valid.

 

I dealt with 3 of them.  Two of my answers were unique to me but

 

The same combination answer that I gave was also given by CPhill DavidQD and probably Alan oh and ofcourse yourself - your answer was correct for this.

 

The 4th interpretation is the most difficult.  Alan and CPhill have both done that one.

Melody  Jan 31, 2015
 #21
avatar+52 
+5

Wow! Everyone's contribution and many thoughts that was put into this question is inspiring! This forum has a wonderful community!  

ultmage11234  Jan 31, 2015
 #22
avatar+92777 
+5

Thank you - that is what we try for here.  A community :)

We want it to be enjoyable and useful for all ages and abilities but we do not want to take ourselves too seriously.  Not all the time anyway :) 

Melody  Jan 31, 2015
 #23
avatar+92777 
+5

A bag contains 2 white b***s,3 black b***s and 4 red b***s. In how many ways can 3 b***s be drawn form the bag, if at least one black ball is to be included?

I think this question is just about dead.

 

If three b***s are chosen at random then

P(at least one is black)

= 1-P(none are black)

$$\\=1- \frac{6C3}{9C3}\\
=1- \frac{20}{84}\\
=\frac{16}{21}\\$$

This is the same as Alan got and the same as DavidQD and CPhill would have gotif carried to its natural conclusions

We did not do it all the same way but this part of our answers are all the same.

-------------

However I do not think that if is correct to say that there are 84 combinations of these 3 b***s or that there are 20 combinations where none are black.  BECAUSE some of these combinations are indistinguishable from each other.

There are only 6 possible outcomes if order does not count.  I think it is correct to call these outcomes combinations  

 

and there are 19 possible outcomes if order does count.  I think it is correct to call these outcomes permutations 

Melody  Feb 2, 2015
 #24
avatar+92777 
+5

A bag contains 2 white b***s,3 black b***s and 4 red b***s. In how many ways can 3 b***s be drawn form the bag, if at least one black ball is to be included?

 

Bertie showed me how to get the permutations for this problem

====================

How many permutations of 3 b***s are there.

Well the first ball can be any of 3 colours, so can the second and so can the third so that is 3*3*3=27  BUT there are only 2 white b***s so one of these (that is WWW) does not exist

so 27-1=26 possible permutations.

====================

how many permutations contain no black one

Well the first one can be W or R,so can the 2nd and the third so 2*2*2 but once again they cannot all be white so we must take 1 away

so 8-1=7 permutations containing no black

=======================

Therefore  26-7=19 permutations DO have at least one black ball.   

=======================

Thanks Bertie.  Some problems are so easy when some smart person shows you how to do them.  

Melody  Feb 3, 2015

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