Combination or probability question

[rajeshmarndi]

Homework Statement

A bag contains 4 black and 5 white balls from which 6 balls are drawn. Determine the number of ways in which at
least 3 black balls can be drawn.

The book has solved it as combination.

Homework Equations

The Attempt at a Solution

When we are drawing say black ball, isn't it the probability of getting 3 black balls at least?
Also will it be correct to number the black ball e.g B1, B2, B3 & B4, which I feel it to be of no sense.

 

[HallsofIvy]

Homework Statement

A bag contains 4 black and 5 white balls from which 6 balls are drawn. Determine the number of ways in which at
least 3 black balls can be drawn.

The book has solved it as combination.

Homework Equations

The Attempt at a Solution

When we are drawing say black ball, isn't it the probability of getting 3 black balls at least?

Sorry, but I have no idea what this means. Isn't what '"the probability of getting 3 black balls at least"?
"At least 3 black balls" means "3 black balls or 4 black balls". Do each of those separately, then add.

Also will it be correct to number the black ball e.g B1, B2, B3 & B4, which I feel it to be of no sense.

You could- and it would make sense- so you could distinguish between the different black balls. But it is not necessary.

Look at 3 black balls first. One way to do that would be to get "BBBWWW" (three black balls first, then three white). How many different ways can "three black and three white balls be drawn? There are 4 possibilities for the first black ball, then 3 for the second and 2 for the third, 5 for the first white ball, 4 for the second and 3 for the third. But then, how many different orders are there for 'BBBWWW'? Then what about 4 black balls, "BBBBWW and different orders?

 

[geoffrey159]

First, you must build a model: you want to find the probability of the event $E = \{ \text{among the 6 balls drawn, at least 3 are black} \}$.
You agree that you can write $E$ as a disjoint union of events $E = E_3 \cup E_4$, where $E_k = \{ \text{among the 6 balls drawn, exactly k of them are black} \}$

The certain event is $\Omega = \{ \text{6 balls are drawn} \}$

Now you know that $P(E) = \frac{\text{card}(E)}{\text{card}(\Omega)}$

What is the cardinal of a disjoint union ?

 

[andrewkirk]

The question is not properly specified. It needs to indicate what is meant by a 'way' in which the ball could be drawn. Here are two different interpretations of 'way' that will give different answers.

1. Before the drawing, number the balls as 1-9. Then a 'way' of drawing 6 balls including exactly 3 blacks is a set of six numbers chosen from 1-9 without replacement, such that exactly three of the numbers denote black balls. Then the question is how many such sets are there. THe interpretation can be further split according to whether the sets are ordered. But since the question mentions 'combination' it probably means not ordered.

2. THe balls are not numbered before the draw but are drawn one at a time. Hence we get a sequence of six 1s and 0s where 1 denotes a black and 0 denotes a white. A 'way' of drawing 6 balls including exactly 3 blacks is a permutation of the sequence 1,1,1,0,0,0. THis one is easy to answer. It's just an unordered drawing of three numbers from 1-6 without replacement.

 

[rajeshmarndi]

The book solution is, as follows.
The possibilities are
(i) 3 black and 3 whites (ii) 4 black and 2 whites

case(i) C(4,3) * C(5,3) = 40 different ways
case(ii) C(4,4) * C(5,2) = 10 different ways.
Hence the total number of ways = 40 + 10 = 50

i.e here the ball are numbered B1, B2 etc i.e B1 B2 B3 W1 W2 W3, B2 B3 B4 W3 W4 W5 etc which doesn't make sense. If we ask how many different ways we can draw balls, it means what are the possible number of pattern of the color, one can get.

So I solve it as follows.

Since we are drawing 6 balls, the possible number of ways black and white balls can be drawn is,

(i) 3 black and 3 white
(ii) 4 black and 2 white

case(i) C(6,3) = 20 , we get the number of ways we see 3 black ball come out when drawn out of 6 attempt. e,g BWBWWB, BBBWWW,...
case(ii) Similarly C(6,4) = 15 , BWWBBB, WWBBBB, ...
Hence the total number of ways = 10 + 15 = 25.

 

[haruspex]

The book solution is, as follows.
The possibilities are
(i) 3 black and 3 whites (ii) 4 black and 2 whites

case(i) C(4,3) * C(5,3) = 40 different ways
case(ii) C(4,4) * C(5,2) = 10 different ways.
Hence the total number of ways = 40 + 10 = 50

i.e here the ball are numbered B1, B2 etc i.e B1 B2 B3 W1 W2 W3, B2 B3 B4 W3 W4 W5 etc which doesn't make sense. If we ask how many different ways we can draw balls, it means what are the possible number of pattern of the color, one can get.

So I solve it as follows.

Since we are drawing 6 balls, the possible number of ways black and white balls can be drawn is,

(i) 3 black and 3 white
(ii) 4 black and 2 white

case(i) C(6,3) = 20 , we get the number of ways we see 3 black ball come out when drawn out of 6 attempt. e,g BWBWWB, BBBWWW,...
case(ii) Similarly C(6,4) = 15 , BWWBBB, WWBBBB, ...
Hence the total number of ways = 10 + 15 = 25.

As andrewkirk noted, both interpretations are reasonable. If I'd had to guess which the book meant, I would have picked correctly. That's because I would assume that they have in mind finding the probability of getting at least 3 black. Note in particular that for the second interpretation you don't need to know how many white balls there are, just that there are at least 3.

 

[rajeshmarndi]

Look at 3 black balls first. One way to do that would be to get "BBBWWW" (three black balls first, then three white). How many different ways can "three black and three white balls be drawn? There are 4 possibilities for the first black ball, then 3 for the second and 2 for the third, 5 for the first white ball, 4 for the second and 3 for the third. But then, how many different orders are there for 'BBBWWW'? Then what about 4 black balls, "BBBBWW and different orders?

According to the question i.e determine the number of ways in which at least 3 black balls can be drawn.

Doesn't it the only thing which makes sense is, what are the possible orders , ball can come out from the bag when drawn. Because same color ball are considered identical.

for e.g, 4 black ball and 2 white ball i.e C(6,4) = 15 ways i.e

if we write the order of black ball coming out in number, it would be as below and remaining is obviously white, and combination of white is hence not required.

1234
1235
1236

1245
1246
1256

1345
1346
1356

1456

2345
2346
2356

2456

3456

Similarly case(ii) 3 black ball and 3 white ball i.e C(6,3) = 20 ways.

Hence the total number of ways = 15 + 20 =35 ways and not 50( book answers).

Is it wrong in doing this way?

 

[haruspex]

According to the question i.e determine the number of ways in which at least 3 black balls can be drawn.

Doesn't it the only thing which makes sense is, what are the possible orders , ball can come out from the bag when drawn. Because same color ball are considered identical.

for e.g, 4 black ball and 2 white ball i.e C(6,4) = 15 ways i.e

if we write the order of black ball coming out in number, it would be as below and remaining is obviously white, and combination of white is hence not required.

1234
1235
1236

1245
1246
1256

1345
1346
1356

1456

2345
2346
2356

2456

3456

Similarly case(ii) 3 black ball and 3 white ball i.e C(6,3) = 20 ways.

Hence the total number of ways = 15 + 20 =35 ways and not 50( book answers).

Is it wrong in doing this way?

I think we ( the people commenting on your post) all agree the question is poorly worded. It admits at least three interpretations. You could answer 2: 3 black and 3 white, or 4 black and 2 white. But you need to understand that the book's interpretation is as valid as yours. To anyone experienced in such problems, the book's interpretation is expected because it relates to the probability of drawing those combinations; your interpretation does not. I am not putting that up as a defence of the book text, it's poor, just explaining how the author came to make this error.

 

[andrewkirk]

I agree with Haruspex. I'll just expand a little on why the fact that the context is probability makes the book's interpretation the expected one. It's because the most common way of measuring probability is to count the number $n$ of equally-likely outcomes that can occur, and then count the number $m$ of those that have the desired characteristic - in this case to have at least three blacks. Basic probability theory then says that the probability is $\frac{m}{n}$.

'Outcome' here means the same as 'way' in the question. It doesn't matter how we define it, as long as all outcomes/ways are equally likely. Under the book's interpretation of numbering the balls before the draw, that requirement is met. Under the interpretation where an outcome/way is a sequence of six characters drawn with replacement from {B,W}, subject to there being no more than four Bs and no more than five Ws, the outcomes/ways are not all equally likely. For instance the outcome WWWWWB is less likely than WWWBBB. So that interpretation is not amenable to a calculation of probability.

 

[geoffrey159]

Take the event $E_k = \{ \text{among the 6 balls, exactly k are black}\}$, $k\le 4$

The elements of $E_k$ are all the couples of $k$ distinct black balls and $6-k$ distinct white balls, so
$E_k = \{ B_{i_1} ... B_{i_k}, \ 1 \le i_1 < ... < i_k \le 4\} \times \{ W_{j_1} ... W_{j_{6-k}}, \ 1 \le j_1 < ... < j_{6-k} \le 5\}$

Just take the definition of the cardinal of a cartesian product and $\text{card}(E_k) = C_4^k C_5^{6-k}$