Solve the given problem involving conditional probability

[chwala]

Homework Statement

see attached (circled question 17).

Relevant Equations

probability

Phew! took time to figure this out...i guess there may be a way to use combinations or markov process i do not know...
anyway,
it was pretty straightforward,

we have the $P_r(w) = \dfrac{n-3}{n}$ from box $X$ and this will result in $P_r(w) = \dfrac{4}{n+1}$ in box $Y$.

Together i shall have $P_r(w)=\dfrac{4(n-3)}{n(n+1)}$

interested in alternative method.

 

[BvU]

we have the $P_r(w) = \dfrac{n-3}{n}$ from box X and this will result in $P_r(w) = \dfrac{4}{n+1}$ in box Y.

So with $n=3$ you have four white balls in box $Y$ ?



$\$

 

[chwala]

So with $n=3$ you have four white balls in box $Y$ ?



$\$

...but should be correct ;

we have $3$ white balls in box $Y$ assuming that 1 white ball is added from box $X$...that sums to $4$.

 

[PeroK]

...but should be correct we have 3 white balls in box $Y$ assuming that 1 white ball is added from box $X$...that sums to $4$.

You should do a probability tree showing all four possible outcomes and their probability.

 

[Delta2]

You need the conditional probability concept here.

The conditions are two
1) The ball taken from box X is black which is a condition C1 with probability P1=... and then the probability that we get white ball from box Y given that the C1 is true is P(white from Y/C1)=...
2) The ball taken from box X is white which is a condition C2 with probability P2=.. and then the probability to get white ball from box Y given that C2 is true P(white from Y/C2)=..

Then you got to combine P1,P(white from Y/C1),P2 and P(white from Y/C2) into an expression that gives the unconditional probability
P(White from Y)=...

 

[BvU]

...but should be correct ;

we have $3$ white balls in box $Y$ assuming that 1 white ball is added from box $X$...that sums to $4$.

With $n=3$ there are only three white balls !!

$\$

 

[Delta2]

@chwala you should also consider the case when the ball taken from box X is black, you still will have some probability for the ball from box Y to be white.

I think btw that the restriction $n\geq 4$ is implied.

 

[chwala]

@chwala you should also consider the case when the ball taken from box X is black, you still will have some probability for the ball from box Y to be white.

I think btw that the restriction $n\geq 4$ is implied.

Let me check on this. Does not look easy I say...

 

[BvU]

Does not look easy I say...

Can you work it out for $n=3$ ?
Then perhaps try n=4 etc until you recognize the pattern

I think btw that the restriction n≥4 is implied.

Why ?

$\$

 

[Delta2]

Let me check on this. Does not look easy I say...

What you did at OP is half the work. The other half work is the case where the ball taken from box X is black. Its pretty much a similar thing. then you add the two probabilities and you have the final probability

Hm i thought the n-3 appears in the denominator somewhere but not. So its ok it can be n=3.

The final probability i get is $$\frac{9+4(n-3)}{n(n+1)}$$

 

[BvU]

We really want to give Chwala a chance to find this for himself....

(and check that it works for $n=3\ \$ ... )

$\$

 

[Delta2]

I worked out b) btw and the answer i get is $$P(X=B/Y=W)=\frac{9}{9+4(n-3)}$$ but I am not sure at all for this.

 

[chwala]

I worked out b) btw and the answer i get is $$P(X=B/Y=W)=\frac{9}{9+4(n-3)}$$ but I am not sure at all for this.

correct as per text book solution; i am looking at part (a) now...

 

[chwala]

@chwala you should also consider the case when the ball taken from box X is black, you still will have some probability for the ball from box Y to be white.

I think btw that the restriction $n\geq 4$ is implied.

I now did this, i can see the pattern but i know that i still need to re read on conditional probability as my problem seems to be with the semantics...anyway as you rightfully indicated $n≠3$.

When $n=4$ we have the probabilities,

$P(w/b)= \dfrac{3}{4}×\dfrac{3}{5} = \dfrac{9}{20}$

$P(w/w)= \dfrac{1}{4}×\dfrac{4}{5} = \dfrac{4}{20}$

$P_{Required} = \dfrac{9}{20} + \dfrac{4}{20} = \dfrac{13}{20}= \dfrac{4n-3}{n(n+1)}$ at $n=4$.

Similarly,

When $n=5$ we shall have

$P_{Required} = \dfrac{9}{30} + \dfrac{8}{20} = \dfrac{17}{30}=\dfrac{4n-3}{n(n+1)}$ at $n=5$.

Now to get the algorithm in term of $n$ for numerator part is what i need to figure out.

 

[Delta2]

The $P_{required}$ for n=4 is indeed 13/20 but where did the $\frac{4n-3}{n(n+1)}$ came from , i thought at post #1 you said $\frac{4(n-3)}{n(n+1)}$ (watch out the parenthesis unless you are based on my post#10 where indeed the numerator is $9+4(n-3)=4n-3$.

 

[Delta2]

Now to get the algorithm in term of n for numerator part is what i need to figure out.

You correctly calculate in post #1 the $P(w/w)=\frac{4(n-3)}{n(n+1)}$, now just calculate the $p(w/b)$ in a very similar way as a function of $n$.

 

[PeroK]

When $n=4$ we have the probabilities, $P(w/b)= \dfrac{3}{4}×\dfrac{3}{5} = \dfrac{9}{20}$

Technically what you have calculated here is $P(X=B)P(Y=W|X=B)$. Or, $P(X=B, Y=W)$.

$P(w/w)= \dfrac{1}{4}×\dfrac{4}{5} = \dfrac{4}{20}$

Likewise, this is $P(X=W, Y=W) = P(X=W)P(Y=W|X=W)$

$P_{Required} = \dfrac{9}{20} + \dfrac{4}{20} = \dfrac{13}{20}= \dfrac{4n-3}{n(n+1)}$ at $n=4$.

$P_{Required}$ is simply $P(Y = W)$. This can be written as:
$$P(Y=W) = P(X=B, Y=W) + P(X=W, Y=W) = P(X=B)P(Y=W|X=B) + P(X=W)P(Y=W|X=W)$$

similarly,
When $n=5$ we shall have $P_{Required} = \dfrac{9}{30} + \dfrac{8}{20} = \dfrac{17}{30}=\dfrac{4n-3}{n(n+1)}$ at $n=5$.

Now to get the algorithm in term of $n$ for numerator part is what i need to figure out.

Okay, but it should be clear what's happening for $n$.

 

[chwala]

Ok i got it we have,

$P(X=B,Y=W) = \dfrac{3}{n} × \dfrac{3}{n+1}= \dfrac{9}{n(n+1)}$

and$P(X=W,Y=W) = \dfrac{n-3}{n} × \dfrac{4}{n+1}= \dfrac{4(n-3)}{n(n+1)}$

adding the two gives us the required value. Cheers!

 

[PeroK]

Part b) is trickier!

 

[chwala]

Part b) is trickier!

Let me look at it...

 

[PeroK]

Let me look at it...

The calculations you have done so far can be shown in a probability tree, which is the best way to tackle these problems. The tree in this case has two initial branches, ecah of which splits into two subbranches:

P(X=B) ($\frac{3}{n}$) --- P(Y=B|X=B) ($\frac{3}{n} \times \frac{n-2}{n+1}$)
---------------- P(Y=W|X=B) ($\frac{3}{n} \times \frac{3}{n+1}$) (**)

P(X=W) ($\frac{n-3}{n}$) --- P(Y=B|X=W) ($\frac{n-3}{n} \times \frac{n-3}{n+1}$)
------------------- P(Y=W|X=W) ($\frac{n-3}{n} \times \frac{4}{n+1}$) (**)

This tells you everything. The two outcomes marked (**) identify a reduced sample space of outcomes where the second ball is white. The outcome you are interested in is the first of these, where the first ball was black.

The probability that the first ball is black given the second ball is white is the probability that the first ball is black relative to the reduced sample space. In other words:
$$P(X=B|Y=W) = \frac{P(Y=W|X=B)}{P(Y=W)} = \frac{P(Y=W|X=B)}{P(Y=W|X=B)+P(Y=W|X=W)}$$

 

[chwala]

The calculations you have done so far can be shown in a probability tree, which is the best way to tackle these problems. The tree in this case has two initial branches, ecah of which splits into two subbranches:

P(X=B) ($\frac{3}{n}$) --- P(Y=B|X=B) ($\frac{3}{n} \times \frac{n-2}{n+1}$)
---------------- P(Y=W|X=B) ($\frac{3}{n} \times \frac{3}{n+1}$) (**)

P(X=W) ($\frac{n-3}{n}$) --- P(Y=B|X=W) ($\frac{n-3}{n} \times \frac{n-3}{n+1}$)
------------------- P(Y=W|X=W) ($\frac{n-3}{n} \times \frac{4}{n+1}$) (**)

This tells you everything. The two outcomes marked (**) identify a reduced sample space of outcomes where the second ball is white. The outcome you are interested in is the first of these, where the first ball was black.

The probability that the first ball is black given the second ball is white is the probability that the first ball is black relative to the reduced sample space. In other words:
$$P(X=B|Y=W) = \frac{P(Y=W|X=B)}{P(Y=W)} = \frac{P(Y=W|X=B)}{P(Y=W|X=B)+P(Y=W|X=W)}$$

I am trying to understand the 'English' on part (b) of the question. The first ball is black... which ball are they reffering to? The ball from box $X$ to $Y$? The second ball ..."is it the drawn ball from box $Y$? if so then i can proceed to analyse. I need to understand the question first.

 

[Delta2]

Yes the first ball is the ball taken from box X and the second ball is the ball taken from box Y (after we have put in Y the ball taken from X).

At least that's how I interpreted it and found the solution presented at post #12

 

[chwala]

Yes the first ball is the ball taken from box X and the second ball is the ball taken from box Y (after we have put in Y the ball taken from X).

At least that's how I interpreted it and found the solution presented at post #12

Ok we have,

$P(B/W) = \dfrac{P(B∩W)}{P(W)} = \dfrac{\dfrac{3}{n}⋅\dfrac{3}{n+1}}{\dfrac{4(n-3)}{n(n+1)}}=\dfrac{9}{n(n+1)}×\dfrac{n(n+1)}{4(n-3)}=\dfrac{9}{4n-3}$

Cheers man!

 

[Delta2]

Not sure if you still fixing Latex but your expression for $P(W)$ is not entirely correct. But yes thats the way to find it.

 

[chwala]

Not sure if you still fixing Latex but your expression for $P(W)$ is not entirely correct. But yes thats the way to find it.

Let me check again...

 

[Delta2]

you say $P(W)=\frac{4(n-3)}{n(n+1)}$ thats not $P(W)$ thats $P(W,B)$

 

[Delta2]

$P(W)$ is the answer to question 1).

 

[chwala]

$P(W)$ is the answer to question 1).

aaargh let me fix that.

 

[chwala]

Ok we have,

$P(B/W) = \dfrac{P(B∩W)}{P(W)} = \dfrac{\dfrac{3}{n}⋅\dfrac{3}{n+1}}{\dfrac{4n-3}{n(n+1)}}=\dfrac{9}{n(n+1)}×\dfrac{n(n+1)}{4n-3}=\dfrac{9}{4n-3}$

Cheers @Delta2

 

[Delta2]

Ehm, not so cheerful here, check your post #18 again, P(W) is the addition of those two.

 

[chwala]

Ehm, not so cheerful here, check your post #18 again, P(W) is the addition of those two.

Yes, i am aware that $P(W)= \dfrac{4n-3}{n(n+1)}$ is the addition given by

$P(X=B,Y=W) = \dfrac{3}{n} × \dfrac{3}{n+1}= \dfrac{9}{n(n+1)}$

and$P(X=W,Y=W) = \dfrac{n-3}{n} × \dfrac{4}{n+1}= \dfrac{4(n-3)}{n(n+1)}$

My post $30$ is correct.

 

[Delta2]

yes ok right, it is $4n-3$ there i thought you had $4(n-3)$.