0
$\begingroup$

We've three jars: jar-$A$ with $1$ red and $1$ white ball, jar-$B$ with $2$ red and $1$ white ball and jar-$C$ with $3$ red balls. We first select a jar and then pick a ball. All selection/picking are random and equally likely. I would like to calculate $\mathbb P($ getting a white ball given jar $A$ is selected $)$.

I define,

  • $A=$ the event that jar-A is selected,

  • $W=$ the event that a white ball is picked, and

I need to find $ℙ(W\mid A)$.

Using concepts of conditional probability, I can focus on a reduced sample space consisting of possible picks from jar A and for this reduced space, can define $W_a=$the event that a white ball is picked. Then $ℙ(W\mid A)$ = $ℙ(W_a)$ = 1/2.

However, I'm stuck when I try to evaluate $ℙ(W\mid A)$ by using the formula, i.e. $ℙ(W∩A)/ℙ(A)$. Put another way, can we evaluate $ℙ(W∩A)$ without using conditional probability?

Note: Implicitly, I've assumed events $A$ and $W$ are defined in the scope of sample space $\Omega=\{\{A,R\}, \{A,W\}, \{B,R1\}, \{B,R2\}, \{B,W\}, \{C,R1\}, \{C,R2\}, \{C,R3\} \} $
and $W_a$ is defined in sample space
$S=\{W,R\}$

$\endgroup$
4
  • $\begingroup$ From $\Omega$ it should be clear that $ℙ(W∩A)=\frac{1}{8}$ or I don't fully understand the question... $\endgroup$
    – Jesús Ros
    Commented Jan 4, 2017 at 12:10
  • $\begingroup$ The way I understand the game...you choose an urn first (probability $\frac 13$ for each) and then choose a ball uniformly from the chosen urn. Assuming I am correct, then the events in your $\Omega $ are not equiprobable (see my post below). $\endgroup$
    – lulu
    Commented Jan 4, 2017 at 12:20
  • $\begingroup$ @gunbl4d3 - The sample points in $\Omega$ are not equi-probable. $\endgroup$
    – KGhatak
    Commented Jan 4, 2017 at 13:55
  • $\begingroup$ @KGhatak true. my mistake. $\endgroup$
    – Jesús Ros
    Commented Jan 4, 2017 at 14:00

1 Answer 1

1
$\begingroup$

Your events are not equiprobable.

Here are the possible events:

I. You choose $A$ and draw $W$, probability $\frac 13\times \frac 12 = \frac 16$.

II. You choose $A$ and draw $R$. Also $\frac 16$

III. You choose $B$ and draw $W$. $\frac 19$

IV. You choose $B$ and draw $R$ $\frac 29$

V. You choose $C$ and draw $W$ prob $0$

VI. You choose $C$ and draw $R$. Prob $\frac 13$.

Consistency check: $\frac 16+\frac 16+\frac 19+\frac 29+\frac 13=1$.

The answer you want is just case I, so $\frac 16$.

$\endgroup$
7
  • $\begingroup$ Your answer is definitely correct. But there are few subtle points that I would like to discuss. To start with, in the sample space of $\Omega$ (which has non-equiprobable sample points), events $A$ and $W$ are not independent. So we cannot write $ℙ(A∩W)=ℙ(A)\times ℙ(W)$ or is it not! $\endgroup$
    – KGhatak
    Commented Jan 4, 2017 at 13:46
  • $\begingroup$ Absolutely correct. Indeed, $P(A)=\frac 13$ and $P(W)=\frac 16+\frac 19=\frac {5}{18}$ $\endgroup$
    – lulu
    Commented Jan 4, 2017 at 14:16
  • $\begingroup$ Great, in that case, would you help putting the case $I$ of your answer in more formal notation. Right now it looks like $ℙ(A∩W) = ℙ(A)\times \frac 12$ and we've just agreed that $ℙ(W)=\frac {5} {18}$ $\endgroup$
    – KGhatak
    Commented Jan 4, 2017 at 14:37
  • $\begingroup$ Not following you. Case I is "choose urn $A$, probability $\frac 13$, then, given that you are in urn $A$, choose $W$, probability $\frac 12$. Not sure what notation you'd have me use. Well, you could write (as you already did) that $P(A\cap W) = P(A) P(W\,|\,A$). But, as I say, you already wrote that. $\endgroup$
    – lulu
    Commented Jan 4, 2017 at 14:40
  • $\begingroup$ I just wrote out all of the possible events explicitly. The formal language is, as you already know, the language of conditional probability. But explicit enumeration is, to me, always the clearest (when it is possible). $\endgroup$
    – lulu
    Commented Jan 4, 2017 at 14:42

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .