Question

From an urn containing 6 balls ,3 white and 3 black ones, a person selects at random an even number of balls (all the different ways of drawing an even number of balls are considered equally probable, irrespective of their number). Tjen the probability that there will be same number of black and white balls among them -

Answer 1

@welshfella

Answer 2

@ganeshie8 how about this problem? :)

Answer 3

I'm assuming that when the question talks about 'an even number of balls' it refers to selecting either 2 or 4 balls, as if we drew all 6 balls from the urn then there would obviously be selecting the same number of black (B) and white (W) balls. So, if this is true then we're looking for:

(Chances of picking the same number of each from 2 balls) OR (Chances of picking the same number of each from 4 balls)

\[=P(Selecting~1W~and~1B~from~2~balls)~+~P(Selecting~2W~and~2B~from~4~balls)\]

For the first case (2 balls), we need to calculate:
\[\frac{ (No.~of~ways~of~selecting~1W~from~3)(No.~of~ways~of~selecting~1B~from~3) }{ Total~No.~of~ways~of~selecting~2~balls~from~6 }\]

For the second case (4 balls), we need to calculate:
\[\frac{ (No.~of~ways~of~selecting~2W~from~3)(No.~of~ways~of~selecting~2B~from~3) }{ Total~No.~of~ways~of~selecting~4~balls~from~6 }\]

So, combining these two, we need to calculate:
\[\frac{ \left(\begin{matrix}3 \\ 1\end{matrix}\right)\left(\begin{matrix}3 \\ 1\end{matrix}\right)}{ \left(\begin{matrix}6 \\ 2\end{matrix}\right) } + \frac{ \left(\begin{matrix}3 \\ 2\end{matrix}\right)\left(\begin{matrix}3 \\ 2\end{matrix}\right) }{ \left(\begin{matrix}6 \\ 4\end{matrix}\right) }\]

\[where~\left(\begin{matrix}3 \\ 1\end{matrix}\right)~means~"3-Choose-1",~for~example\]

The easiest way to work this out is by using the 'choose' function on a scientific calculator, which should give you the answer. Hope that helped! :D

Answer 4

We can simply list all the combinations so that the number of balls is even:
- 3W 3B
3W 1B
- 2W 2B
1W 3B
2W 0B
- 1W 1B
0W 2B

I've written a hyphen in those cases where W = B.

As it is given that all combinations are equally probable, the answer would be \(\frac{3}{7}\).

Answer 5

@Ciarán95 That's pretty close! But your probability comes out to be 1.2 in the end.

Answer 6

Sorry parth bt no option as such..

Answer 7

@ciaran95 i also did that way bt u didn't take the case for drawing out six balls from three black n 3 white balls ..
I included that too in this..
N it came out to be greater than 1 which is indeed impossible..

Answer 8

Could you please tell me what the options are

Answer 9

I mean if you count "zero balls" an option, then the answer is 1/2. But other than that, I'm not sure... I'll have to think it out.

Answer 10

Options are 2/5,4/5,11/15,11/30

Answer 11

This is bothering me a lot.
How're you gonna count the number of ways to select an even number of balls? Have I missed any other cases?

Answer 12

Atleast u r coming up up with an ans which is less than 1 ..
My ans is greater than 1 ...
Imagine my situation ..

Answer 13

U shud be contented with that ...:)

Answer 14

One thing I know is that all balls of the same colour are all supposed to be treated to be the same.

Answer 15

Yes..what's the big deal?

Answer 16

Can you imagine any other way to do this question? It's just really bad...

Answer 17

Will u try to look at his way ??
Ciaran's ..
I did that way only at first ..is it having some fault..?

Answer 18

Except for the ans ...that he ended up with greater than 1

Answer 19

It is.

When he wrote "no. of ways of selecting 1W from 3" as \(\binom 3 1\), it meant that he treated all three balls to be distinct.

Answer 20

So u want to say that it should be just one way to select 1 ball from three balls ..

Answer 21

From Black (white) balls would be more apt

Answer 22

I got 2/5 assuming a particular case of the problem..

Answer 23

@knov

Answer 24

Don't know what is wrong with this ans now?
I came up with ans as 19/31 through other way..

Answer 25

Looking at this again @samigupta8 , I think I may have got this one wrong....I think I misread a really crucial point in the question:

"all the different ways of drawing an even number of balls are considered equally probable, irrespective of their number"

So the way @ParthKohli tackled the question seems to be correct, as all the different possibilities of success occur with the exact same probability when we draw either 2, 4 or 6 balls. So, we just need to pick them our from the list of all possible combinations when we pick 2, 4 or 6 balls. If they didn't occur with the same probability, then we wouldn't be able to do this.

Probability(No. White=No. Black)=Events where No. White=No. BlackTotal No. of Events


So 3/7 appears to be correct in my opinion (Sorry for the confusion!). What method are you using to get 19/31 as an answer?

Answer 26

I used this.
3C1*3C1 + 3C2*3C2 +3C3*3C3/6C2+6C4+6C6

Answer 27

\(\dfrac{\dfrac{ \left(\begin{matrix}3 \\ 1\end{matrix}\right)\left(\begin{matrix}3 \\ 1\end{matrix}\right)}{ \left(\begin{matrix}6 \\ 2\end{matrix}\right) } + \dfrac{ \left(\begin{matrix}3 \\ 2\end{matrix}\right)\left(\begin{matrix}3 \\ 2\end{matrix}\right) }{ \left(\begin{matrix}6 \\ 4\end{matrix}\right) }+1 }{3}\)

I have divided by 3 because all of these 3 probabilities are equally probable.
If they were unequal, then I would have multiplied each probability with the probability of their occurence.

Answer 28

Thanks @hartnn